I’m Up and Down, and Round and Round: Class 9 Ganita Manjari · Chapter 5

Circles, chords, arcs and the angles they make

Ganita Manjari Part 1 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 26 Sep 2026 · Updated 28 Sep 2026

105 min
3 simulations
50 practice questions
36 MCQs

What you will learn in I’m Up and Down, and Round and Round

Describe a circle as a locus and name its centre, radius, chords, diameters and arcs
Explain the rotational and mirror symmetries of a circle and use folding to find a centre
Show why countless circles pass through two points but exactly one passes through three non-collinear points, and construct a circumcircle
Prove and apply the links between equal chords, equal angles at the centre and equal distances from the centre
Use the perpendicular from the centre and the Baudhāyana–Pythagoras theorem to find chord lengths, distances and radii
Prove that an arc makes twice as large an angle at the centre as at the rest of the circle, and use its consequences
Decide when four points are concyclic and use the angle property of cyclic quadrilaterals and its converse

Original companion text aligned to NCERT Mathematics Class 9, Chapter 5, pp. 92–117. Not affiliated with or endorsed by NCERT.

Circles all around us

Look around and circles keep turning up: the rim of a steel katori, a glass bangle, the cut face of an orange, the rings inside a sawn log, the great stone wheels of the Sun Temple at Konark. People have drawn this shape for as long as they have drawn anything.

Why is the circle so common, and why does it look so perfect? It all comes from one simple property. In this chapter we follow that property as far as it goes: how many circles pass through a few given points, how chords behave, and what angle an arc makes at different points. Along the way you will prove each result, not just measure it.

Start with a gardener marking out a round flower bed (Fig. 5.1). She drives a peg into the soil and ties a rope to it. With a stick tied to the other end, she walks round keeping the rope tight, scratching the ground as she goes. Every scratch is exactly one rope-length from the peg, and together the scratches make a circle.

3/3Every point of the curve is 1.5 m from the peg: the peg is the centre.

Fig. 5.1Every point the stick marks is one rope-length from the peg.

A garden bed seen from above, with small flowers in the corners. Step 1 shows a peg in the middle and a rope 1.5 m long stretched to a stick on the right. Step 2 shows the rope swung most of the way round, with the stick having scratched an orange curve about three quarters of the way round the peg. Step 3 shows the finished orange circle; dashed copies of the rope in five other directions show that every point is 1.5 m from the peg, now labelled centre.

ActivityNCERT Activity · p. 93

Circle hunt

You’ll need

a notebook and a pencil
ten minutes and a pair of sharp eyes

Predict first

Which of these is not a circle?

the edge of a one-rupee cointhe outline of a hen's eggthe rim of a bicycle wheela ring painted round a tree trunk, seen from above

Spend ten minutes looking for round things at home, at school or in a park. Aim for at least eight.
Sort them into two groups: made by nature, or made by people.
For each object, point to where you think its centre is.

Table 5.1My circle hunt

Object	Natural or made by people?	Where is its centre?

Your entries are saved automatically.

Warm-up

1 of 2

Any shape that looks round, such as the outline of an egg, is a circle.

TrueFalse

Show answer

Answer False

A circle needs one centre that is equally far from every point of the edge. An egg's outline has no such point, so it is an oval, not a circle.

5.1 Definitions

Every figure in this chapter lives on a flat surface, a plane, such as a sheet of paper or the floor of your classroom. You already know that every point of a circle is the same distance from its centre. Mathematicians go one step further and use that single fact as the definition: nothing else is needed to say exactly what a circle is.

Definition

Circle

A circle (वृत्त) is made of all the points of a plane whose distance from one fixed point of that plane, the centre, equals one fixed length, the radius.

ExampleThe gardener's peg is the centre and her 1.5 m rope is the radius.

Notice the words all the points. The circle is only the boundary curve. The region it encloses is called a disc: a chapati is a disc, and its edge is a circle.

Definition

Locus

A locus (बिंदुपथ) is the set of exactly those points that obey a given rule: every point obeying the rule belongs to it, and no other point does.

ExampleThe locus of points 4 cm from a point $O$ is the circle with centre $O$ and radius 4 cm.

So a circle is the locus of points at a fixed distance from a fixed point. Checking a locus always has two halves: every point on the figure must obey the rule, and every point that obeys the rule must be on the figure. We will need this two-way check in Section 5.3.

Now let us name the parts (Fig. 5.2). Take a circle with centre $O$ . A segment from $O$ to a point $P$ of the circle is a radius (त्रिज्या); its length $r$ is also called the radius. A segment joining two points $A$ and $B$ of the circle is a chord (जीवा).

Join $A$ and $B$ to the centre. The angle $∠ A O B$ is the angle that chord $A B$ subtends at the centre, the opening you see when you look at the chord from $O$ . A chord through the centre, such as $C D$ , is a diameter (व्यास). It is made of two radii, so its length is $2 r$ .

4/4A chord through the centre is a diameter, of length $2 r$ .

Fig. 5.2The parts of a circle that we use in every theorem.NCERT Fig. 5.3 · p. 93

A circle with centre O. Step 1 shows the centre O and a green radius OP of length r. Step 2 adds a blue chord AB near the top of the circle. Step 3 joins A and B to O with grey radii and shades the angle AOB that the chord subtends at the centre. Step 4 adds an orange diameter CD through O, labelled diameter equals 2r.

Quick check

1 of 4

A calf is tied to a pole with a 3 m rope and walks round keeping the rope tight. Which figure do its positions make?

a filled disc of radius 3 ma circle of diameter 3 ma circle of radius 3 m centred at the polea square of side 3 m

Show answer

Answer (C) a circle of radius 3 m centred at the pole

With the rope tight, the calf is always exactly 3 m from the pole, so it stays on the circle of radius 3 m. The disc is the grazing area when the rope may go slack.

5.2 Symmetries of a Circle

Part of what makes a circle so pleasing is its perfect symmetry. Put a plain steel thali on a table and close your eyes while a friend turns it. When you open them, can you tell how far it was turned? You cannot. The plate looks exactly the same after a turn through any angle about its centre.

A square tile would give the game away: it looks unchanged only after a quarter turn, a half turn or a three-quarter turn. We say that a circle has complete rotational symmetry about its centre (Fig. 5.3).

A circle also has mirror symmetry, and far more of it than a square. A square has only four mirror lines. For a circle, every diameter works: the two halves on either side of it are mirror images. Paper shows this quickly. Any fold that lays the edge of a disc exactly on itself creases along a diameter, and you may choose its direction freely.

4/4Every diameter of a circle is a line of symmetry.

Fig. 5.3A circle looks the same after any turn, and it can be folded along any diameter.

A blue square and an orange circle side by side, each carrying a small red sticker. Step 1 shows both at rest. Step 2 turns both by 30 degrees: the square visibly tilts and looks different, while the circle looks the same except that its sticker has moved. Step 3 shows the four dashed mirror lines of the square and one dashed diameter of the circle as a fold line. Step 4 shows many dashed diameters across the circle, each of them a mirror line.

Try this

Find the centre by folding

You’ll need

a paper disc (trace round a steel plate or a katori and cut it out)
a pencil and a ruler

Predict first

Two creases, each folded so that the edge lies on itself, cross at one point. What is that point?

the centre of the disca point on the edgeit depends on which folds you make

Fold the disc so that its edge lies exactly on itself. Press to make a crease, then open it out.
Fold it again in a different direction in the same way, and open it.
Mark the point where the two creases cross. Measure from it to the edge in several directions.

5.3 How Many Circles?

Mark two points $A$ and $B$ on a page. How many circles pass through both of them?

The smallest circle through $A$ and $B$ is easy to find. Put its centre at the midpoint $M$ of $A B$ . Since $M A = M B$ , the circle with centre $M$ and radius $M A$ passes through both points, and $A B$ is one of its diameters, so its radius is $\frac{1}{2} A B$ . No circle through $A$ and $B$ can be smaller: $A B$ is a chord of every such circle, and no chord is longer than a diameter (we prove this in Section 5.6.1).

Are there others? The centre $O$ of any circle through $A$ and $B$ must have $O A = O B$ , since both are radii. So we need to know all the points that are equally far from $A$ and $B$ .

LemmaThe perpendicular bisector as a locus

The points that are equally far from two points $A$ and $B$ are exactly the points of the perpendicular bisector of $A B$ .

Mark the midpoint $M$ of $A B$ , and let $ℓ$ be the line through $M$ at right angles to $A B$ . We make the two-way check.

Every point of $ℓ$ is equally far from $A$ and $B$ . Take a point $P$ on $ℓ$ other than $M$ (for $M$ itself it is clear).

P A = P B

Every point equally far from $A$ and $B$ lies on $ℓ$ . Take a point $Q$ with $Q A = Q B$ that is not on line $A B$ (the only such point on $A B$ is $M$ ).

Q M ⊥ A B

Both halves hold, so the perpendicular bisector is exactly the locus.

So the possible centres of circles through $A$ and $B$ are exactly the points of the perpendicular bisector of $A B$ , and there are infinitely many of them (Fig. 5.4). As the centre slides away from $A B$ along the bisector, the radius grows and the circle looks flatter near $A$ and $B$ .

5/5Every centre is on the bisector, and every point of the bisector is a centre.

Fig. 5.4Every point of the perpendicular bisector of

A B

is the centre of a circle through

A

and

B

.NCERT Fig. 5.4 · p. 95

Points A and B, with segment AB slanting up to the right, and the dashed blue perpendicular bisector of AB crossing it at right angles. Step 1 shows only A, B, the equal-half marks and the bisector. Step 2 adds the smallest circle, centred at the midpoint M, with AB as a diameter. Step 3 shows a bigger circle centred at P, further along the bisector. Step 4 shows a circle centred at Q, on the other side of AB. Step 5 shows all of these circles together with a very large one centred at R; every centre lies on the bisector.

Now take three points $A$ , $B$ and $C$ . If they are collinear (संरेख), lying on one straight line, then no circle passes through all three. The perpendicular bisectors of $A B$ and $B C$ are both perpendicular to that line, so they are parallel and never meet. No point is equally far from all three. This also shows that a straight line meets a circle in at most two points.

If the three points are not collinear, the story changes completely.

Theorem 1Circle through three points

Through any three points that do not lie on one straight line, there passes exactly one circle.

Let the points be $A$ , $B$ and $C$ , not on one line (Fig. 5.5). Suppose a circle passes through all three, with centre $O$ . We work out where $O$ must be.

O = the crossing point of two perpendicular bisectors

So there is only one possible centre. And it works: the crossing point is equally far from $A$ and $B$ , and from $A$ and $C$ , so the circle with that centre and radius $O A$ passes through all three points. One centre and one radius give exactly one circle.

4/4They cross at $O$ , equally far from all three points. The third bisector passes through $O$ too.

Fig. 5.5Two perpendicular bisectors locate the one and only centre.NCERT Fig. 5.5 · p. 96

Three points A, B and C joined by a dashed triangle. Step 1 shows only the points. Step 2 draws the blue perpendicular bisector of AB, with equal-half marks and a right angle. Step 3 adds the purple perpendicular bisector of AC, which crosses the first line. Step 4 marks the crossing point O, draws the orange circle through A, B and C with equal radii, and shows that the dashed green bisector of BC passes through O as well.

Definition

Circumcircle and circumcentre

The circle through the three vertices of a triangle is its circumcircle (परिवृत्त), and its centre is the circumcentre (परिकेंद्र). We say the circle circumscribes the triangle, and the triangle is inscribed in the circle.

ExampleBy Theorem 1, every triangle has exactly one circumcircle.

The crossing point $O$ is also equally far from $B$ and $C$ , so it lies on the third perpendicular bisector as well. The three perpendicular bisectors of a triangle always meet at one point, the circumcentre. Where that point sits depends on the shape of the triangle (Fig. 5.6).

2/3 $∠ A = 90^{\circ}$ : $O$ is the midpoint of the hypotenuse $B C$ .

Fig. 5.6Where the circumcentre sits tells you what kind of triangle it is.NCERT Figs. 5.5–5.7 · p. 97

Side BC stays fixed while vertex A moves; each step shows the blue triangle, its orange circumcircle and the circumcentre O with dashed radii. Step 1: an acute triangle, with O inside it. Step 2: angle A is a right angle and O sits exactly at the midpoint of BC, the hypotenuse. Step 3: angle A is obtuse and O lies below BC, outside the triangle.

The right-angled case is worth remembering. Its circumcentre is the midpoint of the hypotenuse, so the hypotenuse is a diameter of the circumcircle. You will see why in Section 5.7.1.

ActivityNCERT Exercise Set 5.1 · p. 98

Construct a circumcircle

You’ll need

a ruler
a pair of compasses
a protractor
a sharp pencil

Predict first

A triangle has angles

40^{\circ}

30^{\circ}

and

110^{\circ}

. Where will its circumcentre be?

inside the triangleon one of its sidesoutside the triangle

Draw any triangle $P Q R$ with sides between 5 cm and 8 cm.
Open the compasses to more than half of $P Q$ . Draw arcs from $P$ and from $Q$ on both sides of $P Q$ , and join the two crossing points. This line is the perpendicular bisector of $P Q$ .
Repeat for side $Q R$ . Mark the point $O$ where the two bisectors meet.
Put the compass point on $O$ , open it to $P$ and draw the circle. Does it pass through $Q$ and $R$ ?
Measure $O P$ , $O Q$ and $O R$ , and note whether $O$ is inside the triangle.

Table 5.2Circumcircle measurements

Triangle	OP (cm)	OQ (cm)	OR (cm)	O inside?
My triangle
A friend's triangle

Your entries are saved automatically.

Quick check

1 of 4

Two points

A

and

B

are 10 cm apart. What is the smallest possible radius of a circle through both?

10 cm5 cm2.5 cmthere is no smallest radius

Show answer

Answer (B) 5 cm

The smallest circle has

A B

as a diameter, so its radius is

10 \div 2 = 5

cm. There is no largest circle, but there is a smallest one.

Practice · 6 questions

Practice 5.1: questions with answers

DiagramEasyNCERT Exercise Set 5.1, Q1
Construct $△ P Q R$ with $Q R = 6$ cm, $∠ Q = 50^{\circ}$ and $∠ R = 65^{\circ}$ . Draw its circumcircle. Is the circumcentre inside or outside the triangle?
Show solution
AnswerInside, because all three angles ( $65^{\circ}$ , $50^{\circ}$ , $65^{\circ}$ ) are acute.
Draw $Q R = 6$ cm, then angles of $50^{\circ}$ at $Q$ and $65^{\circ}$ at $R$ . Their arms meet at $P$ .
$∠ P = 180^{\circ} - 50^{\circ} - 65^{\circ} = 65^{\circ}$ , so every angle is acute.
Draw the perpendicular bisectors of $Q R$ and $P Q$ ; they meet at $O$ . The circle with centre $O$ and radius $O Q$ passes through $P$ , $Q$ and $R$ , and $O$ lies inside the triangle.
DiagramEasyNCERT Exercise Set 5.1, Q2
Construct $△ P Q R$ with $P Q = 4.5$ cm, $∠ P = 115^{\circ}$ and $P R = 5$ cm. Draw its circumcircle. Is the circumcentre inside or outside the triangle?
Show solution
AnswerOutside, beyond side $Q R$ , because $∠ P$ is obtuse.
Draw $P Q = 4.5$ cm, make $∠ Q P R = 115^{\circ}$ at $P$ and mark $R$ with $P R = 5$ cm. Join $Q R$ .
Draw the perpendicular bisectors of $P Q$ and $P R$ . They meet at a point $O$ on the far side of $Q R$ from $P$ .
The circle with centre $O$ through $P$ also passes through $Q$ and $R$ . Since $∠ P$ is obtuse, $O$ is outside the triangle.
DiagramMediumNCERT Exercise Set 5.1, Q3
Construct $△ X Y Z$ with $X Y = 5$ cm, $Y Z = 6$ cm and $Z X = 7$ cm. Draw its circumcircle, mark the circumcentre $O$ and measure $O X$ , $O Y$ and $O Z$ .
Show solution
AnswerAll three measure about 3.6 cm.
Construct the triangle from its three sides, draw the perpendicular bisectors of two sides and mark their crossing point $O$ .
$O X$ , $O Y$ and $O Z$ all measure about $3.6$ cm (a careful calculation gives $3.57$ cm). They are equal because $O$ lies on the perpendicular bisectors, so it is equally far from all three vertices.
NumericalEasyNCERT Exercise Set 5.1, Q4
Two points are 7 cm apart. How small can the radius of a circle through both of them be?
cm
Show answer
Answer 3.5 cm
The smallest such circle has the segment as a diameter, so its radius is $7 \div 2 = 3.5$ cm.
5
ProofMediumNCERT Ex. Set 5.1, Think, Draw and Infer Q1
Points $P$ , $Q$ and $R$ lie on one line, with $P Q = 3$ cm and $Q R = 4$ cm.
1. Draw the perpendicular bisectors of $P Q$ and $Q R$ . How are they related?
  Show solution
  AnswerThey are parallel, 3.5 cm apart.
  Both bisectors are perpendicular to line $P R$ , so they are parallel. They cross $P R$ at the midpoints of $P Q$ and $Q R$ , which are $1.5 + 2 = 3.5$ cm apart.
2. Is there a point that is equally far from $P$ , $Q$ and $R$ ?
  Show solution
  AnswerNo.
  Such a point would have to lie on both bisectors, but parallel lines never meet.
3. Can a straight line cut a circle in three different points?
  Show solution
  AnswerNo.
  If it could, those three collinear points would all lie on one circle, which part (b) rules out.
Short answerMediumNCERT Ex. Set 5.1, Think, Draw and Infer Q2
The circumcircle of $△ A B C$ is drawn. Can a different triangle, congruent to $△ A B C$ , have the same circumcircle? If so, how many such triangles are there?
Show solution
AnswerYes, infinitely many.
Turn $△ A B C$ about the circumcentre $O$ through any angle. Each vertex stays the same distance from $O$ , so the turned triangle is congruent to $△ A B C$ and its vertices are still on the circle.
Reflecting in any diameter works too. There are infinitely many angles to turn through, so there are infinitely many such triangles.

5.4 Chords and the Angles They Subtend

Every chord comes with an angle: the angle it subtends at the centre. Long chords seem to go with wide angles and short chords with narrow ones. In this section we make that link exact.

Look at a wall clock. The straight line from the 12 mark to the 2 mark is a chord of its rim, and so is the line from the 3 mark to the 5 mark. A quarter turn of the clock face carries the first chord exactly onto the second. So the two chords have the same length, and both make the same angle, $60^{\circ}$ , at the centre. A turn about the centre does this to any chord (Fig. 5.7).

3/3 $A B = C D$ and $∠ A O B = ∠ C O D$ .

Fig. 5.7Turning a chord about the centre keeps both its length and its angle at the centre.NCERT Figs. 5.8–5.10 · p. 98

A circle with centre O. Step 1 shows a blue chord AB near the lower left, with radii OA and OB and the 80 degree angle between them. Step 2 turns the chord, its radii and its angle about O through 150 degrees to a new position CD near the top, leaving a dashed purple copy of AB behind. Step 3 shades the triangles OAB and OCD, marks AB and CD as equal, and shows the 80 degree angle at the centre for both.

Turning makes the result believable, but a proof must work for any two equal chords, including ones we did not get by turning. Congruent triangles do the job.

Theorem 2Equal chords, equal angles

If two chords of a circle are equal in length, the angles they subtend at the centre are equal.

Now turn the question around. Suppose two chords make the same angle at the centre. Must they be equal? Think of a hand fan pinned at its centre. Open two of its ribs to a fixed angle and spin the fan: the tips of the two ribs always stay the same distance apart.

Theorem 3Equal angles, equal chords

When two chords of one circle make equal angles at its centre, the two chords have the same length.

Example 5.1

Angles from equal chords

In a circle with centre $O$ , chords $P Q$ and $R S$ are equal and $∠ P O Q = 48^{\circ}$ . Find $∠ R O S$ and $∠ O R S$ .

Try it yourself first. Then reveal the solution one step at a time.

Practice · 2 questions

Practice 5.2: questions with answers

1
ProofEasyNCERT Exercise Set 5.2, Q1
Chord $A B$ of a circle with centre $O$ subtends $∠ A O B = 110^{\circ}$ .
1. Explain why $△ O A B$ is isosceles.
  Show solution
  Answer $O A = O B$ .
  $O A$ and $O B$ are both radii of the circle, so they are equal. A triangle with two equal sides is isosceles.
2. Find $∠ O A B$ in degrees.
  °
  Show answer
  Answer 35 °
  $∠ O A B = ∠ O B A = (180^{\circ} - 110^{\circ}) \div 2 = 35^{\circ}$ .
ProofMediumNCERT Exercise Set 5.2, Q2
$A B$ and $C D$ are equal chords of a circle with centre $O$ . Show that $△ O A B ≅ △ O C D$ , and hence that these two triangles have equal areas.
Show solution
$O A = O C$ and $O B = O D$ (radii), and $A B = C D$ (given). By the SSS rule, $△ O A B ≅ △ O C D$ .
Congruent triangles fit exactly on top of each other, so they have equal areas.

5.5 Midpoints and Perpendicular Bisectors of Chords

Draw a chord, mark its midpoint and join the midpoint to the centre. What angle do you get? And what happens if we start instead from the perpendicular dropped from the centre onto the chord?

Theorem 4Centre to midpoint

If $M$ is the midpoint of a chord $A B$ (not a diameter) of a circle with centre $O$ , then $O M$ is perpendicular to $A B$ .

5/5Two equal angles on a straight line: each is $90^{\circ}$ .

Fig. 5.8Two congruent halves force a right angle at the midpoint.NCERT Fig. 5.12 · p. 101

A circle with centre O and a blue chord AB below the centre. Step 1 shows only the chord. Step 2 marks its midpoint M with equal-half marks. Step 3 draws the orange segment OM and the grey radii OA and OB. Step 4 shades triangles OMA and OMB in two colours with matching marks and lists OA = OB, AM = MB and OM common, so SSS. Step 5 adds a right-angle mark at M: the two equal angles add up to 180 degrees, so each is 90 degrees.

The converse is true as well, and this time the RHS rule does the work.

Theorem 5Perpendicular from the centre

A perpendicular dropped from the centre of a circle onto a chord divides that chord into two equal parts.

Together the two theorems give a very handy fact: the perpendicular bisector of every chord passes through the centre. The perpendicular from $O$ halves the chord, so it is the chord's perpendicular bisector. Draw two chords and bisect both: the bisectors cross at the centre.

Theorem 5 also turns chord problems into right-triangle problems. The radius $O B$ , the distance $O M$ and the half-chord $M B$ form a triangle with its right angle at $M$ , so the Baudhāyana–Pythagoras theorem applies.

Half-chord relation

r^{2} = d^{2} + {(\frac{c}{2})}^{2}

where

$r$: the radius
$d$: the distance of the chord from the centre
$c$: the length of the chord

Example 5.2

A plank across a well

The mouth of a circular well has radius 1.3 m. A straight plank lies across the mouth, and its edge forms a chord 0.5 m from the centre. How long is this chord?

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 4

O

is the centre of a circle and

M

is the midpoint of a chord

A B

that is not a diameter. Which statement is always true?

O M = A M

O M = \frac{1}{2} r

O M ⊥ A B

M

lies on the circle

Show answer

Answer (C) $O M ⊥ A B$

By Theorem 4, the segment from the centre to the midpoint is perpendicular to the chord.

O M = A M

happens only for one particular chord.

Practice · 3 questions

Practice 5.3: questions with answers

ProofMediumNCERT Exercise Set 5.3, Q1
Give a second proof of Theorem 5, using the Baudhāyana–Pythagoras theorem instead of congruent triangles.
Show solution
Since $O M ⊥ A B$ , triangles $O M A$ and $O M B$ are right-angled at $M$ .
$A M^{2} = O A^{2} - O M^{2} = O B^{2} - O M^{2} = M B^{2}$
This uses $O A = O B$ (radii). Lengths are positive, so $A M = M B$ .
ProofMediumNCERT Exercise Set 5.3, Q2
Triangle $P Q R$ is inscribed in a circle with centre $O$ , and $P Q = P R$ . Show that the line joining $P$ to the midpoint $N$ of $Q R$ passes through $O$ .
Show solution
$P Q = P R$ , so $P$ is equally far from $Q$ and $R$ and lies on the perpendicular bisector of $Q R$ .
$O$ is the centre, so $O Q = O R$ , and $O$ lies on the same bisector. That bisector also passes through $N$ , the midpoint of $Q R$ .
So $P$ , $N$ and $O$ all lie on one line, the perpendicular bisector of $Q R$ .
NumericalMediumNCERT Exercise Set 5.3, Q3
Two parallel chords of a circle of radius 17 cm are 16 cm and 30 cm long, and they lie on opposite sides of the centre. How far apart are their midpoints?
cm
Show answer
Answer 23 cm
The distances are $\sqrt{17^{2} - 8^{2}} = 15$ cm and $\sqrt{17^{2} - 15^{2}} = 8$ cm. Both midpoints lie on the diameter perpendicular to the chords, on opposite sides of $O$ , so they are $15 + 8 = 23$ cm apart.

5.6 Distance of Chords from the Centre

The distance from a point to a line means the length of the perpendicular from the point to the line, the shortest path between them. So when we ask how far a chord is from the centre, we mean the length $O M$ , where $M$ is the foot of the perpendicular from $O$ . By Theorem 5, $M$ is also the midpoint of the chord.

4/4The creases meet at $M$ at $90^{\circ}$ , and $O M$ is the distance of $A B$ from the centre.

Fig. 5.9Two folds make a chord and the perpendicular to it from the centre.NCERT Fig. 5.13 · p. 102

A pale paper disc with its centre marked. Step 1 shows the plain disc. Step 2 shows a flap folded over along a dashed blue crease, which is the chord AB. Step 3 shows a second fold that brings A onto B, with a purple crease through the centre and the folded half shaded. Step 4 shows both creases opened out: they meet at M at a right angle, M halves AB, and the orange segment OM is the distance of the chord from the centre.

ActivityNCERT Activity · p. 102

Fold, crease and measure

You’ll need

a paper disc with its centre marked
a ruler and a protractor
tracing paper or butter paper
a pair of compasses

Predict first

Two chords of the same circle have the same length. How will their distances from the centre compare?

the chord drawn first will be closerthey will be equalit depends on where the chords are

Fold a flap of the disc inwards and crease it. Open it: the crease is a chord $A B$ .
Fold the disc so that $A$ lands exactly on $B$ , crease and open. Call the point where the creases cross $M$ .
Measure $A M$ , $M B$ , the angle between the creases, and $O M$ .
Set the compasses to the length $A B$ . Put the point anywhere on the edge and mark where the pencil meets the edge again: that gives a second chord of the same length. Measure its distance from $O$ as well.
Trace the circle, the first chord and $O M$ onto tracing paper. Keep the two centres together and turn the tracing: the traced chord lies over other chords of the same length, and the traced $O M$ turns with it.

Table 5.3Two equal chords

Chord	Length (cm)	Distance from centre (cm)
First chord
Second chord

Your entries are saved automatically.

Can’t do it now? Try the simulation · Fig. 5.11

Measuring a few chords, however carefully, tells us only about those chords. To be sure the pattern never fails, we need a reason that covers every pair of equal chords at once. Right triangles supply it.

Theorem 6Equal chords are equally far from the centre

Equal chords of a circle lie at equal distances from its centre.

Let $A B$ and $C D$ be equal chords of a circle with centre $O$ , with $O M ⊥ A B$ and $O N ⊥ C D$ . By Theorem 5, $M$ and $N$ are midpoints, so $A M = \frac{1}{2} A B = \frac{1}{2} C D = C N$ .

Fig. 5.10Right triangles

O M A

and

O N C

are congruent (RHS).NCERT Fig. 5.14 · p. 103

A circle with centre O and two equal chords, a blue chord AB at the lower left and a purple chord CD at the upper right. Orange perpendiculars OM and ON meet the chords at right angles at their midpoints M and N, and dashed radii OA and OC complete the right triangles OMA and ONC.

O M = O N

A second route: triangles $O A B$ and $O C D$ are congruent by SSS, and congruent triangles have equal heights on matching sides, so $O M = O N$ .

Theorem 7Chords equally far from the centre are equal

If two chords of a circle are the same distance from the centre, they have the same length.

SimulationTwo chords in one circle

A circle with centre O and two chords, AB in blue and CD in purple, whose four endpoints you can drag round the circle. Perpendiculars OM and ON show each chord's distance from the centre, and a panel lists the two chord lengths, the two distances and the two angles at the centre. A button makes CD equal to AB, and a message says how the lengths and the distances compare.

Fig. 5.11Drag the four points. The panel shows each chord's length, its distance from the centre and its angle at the centre.

Try this

Press Make CD = AB. Compare $O M$ with $O N$ , and $∠ A O B$ with $∠ C O D$ .
Drag $D$ so that $C D$ is longer than $A B$ . Which chord is now nearer to $O$ ?
Try to make a chord longer than a diameter. What stops you?

Example 5.3

How far is each chord?

Two chords of a circle of radius 6.5 cm are each 12 cm long. How far is each chord from the centre?

Try it yourself first. Then reveal the solution one step at a time.

Practice · 3 questions

Practice 5.4: questions with answers

ProofMediumNCERT Exercise Set 5.4, Q1
Prove Theorem 6 again, this time with the Baudhāyana–Pythagoras theorem instead of congruent triangles.
Show solution
$O M^{2} = r^{2} - A M^{2}, O N^{2} = r^{2} - C N^{2}$
Since $A B = C D$ , the half-chords are equal: $A M = C N$ . So $O M^{2} = O N^{2}$ , and because both distances are positive, $O M = O N$ .
ProofMediumNCERT Exercise Set 5.4, Q2
In a circle with centre $O$ , $A B$ and $C D$ are chords with $O M ⊥ A B$ , $O N ⊥ C D$ and $O M = O N$ . Use congruent triangles to prove that $A B = C D$ .
Show solution
Triangles $O M A$ and $O N C$ are right-angled at $M$ and $N$ , with hypotenuses $O A = O C$ (radii) and legs $O M = O N$ (given). By RHS they are congruent, so $A M = C N$ .
By Theorem 5, $A B = 2 A M$ and $C D = 2 C N$ . Hence $A B = C D$ .
3
NumericalMediumNCERT Exercise Set 5.4, Q3
Chords $A B$ and $C D$ of a circle of radius 5 cm are each 4 cm from the centre.
1. How long is each chord, in cm?
  cm
  Show answer
  Answer 6 cm
  Half-chord $= \sqrt{5^{2} - 4^{2}} = 3$ cm, so each chord is $6$ cm.
2. Use the Baudhāyana–Pythagoras theorem to explain why two chords at the same distance $d$ from the centre of any circle must be equal.
  Show solution
  Each half-chord equals $\sqrt{r^{2} - d^{2}}$ , which depends only on $r$ and $d$ . Two chords at the same distance therefore have equal halves, and so equal lengths.

5.6.1 Which of the two unequal chords is farther from the centre?

Equal chords sit at equal distances from the centre. What about unequal chords: is the longer one nearer the centre or farther out? Decide on your answer now, then test it in the activity below.

ActivityNCERT Activity · p. 104

Chord survey

You’ll need

a pair of compasses
a ruler
a set square

Predict first

In one circle you draw chords 2 cm, 6 cm and 10 cm long. Which will be nearest to the centre?

the 2 cm chordthe 6 cm chordthe 10 cm chordall equally near

Draw a circle of radius 5 cm and mark its centre $O$ .
Set the compasses to 2 cm, put the point on the circle and mark where the pencil crosses the circle. Join the two points to get a 2 cm chord. Draw chords of 4, 6, 8 and 10 cm in the same way, spread round the circle.
From $O$ , drop a perpendicular to each chord using a set square, and measure it.
Record your results in the table.

Table 5.4Chord length and distance from the centre

Chord length	Distance from O (cm)
2 cm
4 cm
6 cm
8 cm
10 cm

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5/5Longer chords lie nearer to the centre.

Fig. 5.12As a chord moves away from the centre, it gets shorter.NCERT Fig. 5.16 · p. 105

A circle of radius 5 cm with a horizontal chord below the centre O. Step 1: the chord passes through O, so its distance d is 0 and it is a 10 cm diameter. Step 2: the chord is 3 cm below O and 8 cm long, with the right triangle formed by the distance, half the chord and a radius. Step 3: 4 cm below O, the chord is 6 cm long. Step 4: 4.8 cm below O, it is only 2.8 cm long. Step 5 shows all four chords together with a table pairing each length with its distance.

Theorem 8The longer chord is nearer

Of two unequal chords of a circle, the longer one is nearer to the centre.

Let $A B$ and $C D$ be chords of a circle with centre $O$ and radius $r$ , with $A B > C D$ . Drop perpendiculars $O M$ onto $A B$ and $O N$ onto $C D$ . By Theorem 5, $M$ and $N$ are midpoints, so $A M = \frac{1}{2} A B$ and $C N = \frac{1}{2} C D$ .

O M < O N

Theorem 8 settles the question from Section 5.2 about the longest chord. A distance can never be less than $0$ , and only a chord through the centre is at distance $0$ . That chord is a diameter, so it is nearer the centre than any other chord, and therefore no chord is longer than a diameter. At the other extreme, a chord drawn very close to the edge is very short, as the last steps of Fig. 5.12 show.

Example 5.4

A formula for the length of a chord

Show that a chord at distance $d$ from the centre of a circle of radius $r$ has length $2 \sqrt{r^{2} - d^{2}}$ . Use this to find the length of a chord 4 cm from the centre of a circle of radius 8.5 cm.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 4

In one circle, chord

P Q

is 7 cm long and chord

R S

is 9 cm long. Which is nearer to the centre?

P Q

R S

they are equally nearit cannot be decided

Show answer

Answer (B) $R S$

By Theorem 8, the longer chord is nearer. Many people guess the opposite, picturing a longer chord as bigger and so further out.

Practice · 3 questions

Practice 5.5: questions with answers

NumericalEasyNCERT Exercise Set 5.5, Q1
The radius of a circle is 6.1 cm and a chord is 1.1 cm from the centre. How long is the chord?
cm
Show answer
Answer 12 cm
Half-chord $= \sqrt{{6.1}^{2} - {1.1}^{2}} = \sqrt{37.21 - 1.21} = \sqrt{36} = 6$ cm, so the chord is $12$ cm long.
2
ProofMediumNCERT Exercise Set 5.5, Q2
The chord-length formula from Example 5.4 is $c = 2 \sqrt{r^{2} - d^{2}}$ .
1. Rearrange it to give the distance $d$ in terms of $r$ and $c$ , and say what each step means in the figure.
  Show solution
  Answer $d = \sqrt{r^{2} - \frac{c^{2}}{4}}$
  $\frac{c}{2} = \sqrt{r^{2} - d^{2}}$ , so $\frac{c^{2}}{4} = r^{2} - d^{2}$ and $d = \sqrt{r^{2} - \frac{c^{2}}{4}}$ .
  It is the same right triangle $O M B$ read the other way round: the distance is one leg, half the chord is the other leg, and the radius is the hypotenuse.
2. How far is a 5 cm chord from the centre of a circle of radius 6.5 cm?
  cm
  Show answer
  Answer 6 cm
  $d = \sqrt{{6.5}^{2} - {2.5}^{2}} = \sqrt{42.25 - 6.25} = \sqrt{36} = 6$ cm.
ProofHardNCERT Exercise Set 5.5, Q3
In a circle of radius 5 cm, chord $P Q$ is 4 cm from the centre and chord $R S$ is 2 cm from the centre, so $P Q$ is twice as far away. Is $R S$ twice as long as $P Q$ ? Can we say in general that doubling the distance halves the chord?
Show solution
AnswerNo: $P Q = 6$ cm but $R S \approx 9.17$ cm, not 12 cm.
$P Q = 2 \sqrt{25 - 16} = 6$ cm and $R S = 2 \sqrt{25 - 4} = 2 \sqrt{21} \approx 9.17$ cm, so $R S$ is not $2 \times P Q$ .
In general we cannot conclude it: the length depends on $\sqrt{r^{2} - d^{2}}$ , not on $d$ in simple proportion. It can happen for one special pair of chords (when the nearer chord is at distance $r / \sqrt{5}$ ), so the claim is not always false, but it is certainly not always true.

5.7 Angles Subtended by an Arc

Two points $A$ and $B$ on a circle cut it into two pieces, and each piece is an arc (चाप) from $A$ to $B$ . The shorter piece is the minor arc and the longer one the major arc. Both arcs have the same end points, so we name an arc by adding a point that lies on it: in Fig. 5.13, $A S B$ is the minor arc and $A T B$ the major arc. When $A B$ is a diameter, the two pieces are equal and each is a semicircle.

Definition

Arc

An arc is the part of a circle you trace when you move along it from one point to another without lifting your pencil. The two points are its end points, and they always split the circle into two arcs: a minor arc and a major arc, or two semicircles.

ExampleOn a clock face, the rim from 12 to 3 (going clockwise) is a minor arc; the rim from 3 on round to 12 is a major arc.

To measure the angle an arc subtends at the centre, picture a radius that starts along $O A$ and turns about $O$ , its tip running along the arc, until it lies along $O B$ . The amount it turns is the angle. For the minor arc this is the ordinary angle $∠ A O B$ , less than $180^{\circ}$ . For the major arc the radius goes the long way round, so the angle is reflex, more than $180^{\circ}$ . The two angles together make a full turn of $360^{\circ}$ .

4/4The two angles at the centre add up to $360^{\circ}$ .

Fig. 5.13Every pair of points splits a circle into a minor arc and a major arc.NCERT Figs. 5.17–5.18 · p. 106

A circle with centre O and two points A and B on its right-hand side, with S on the short arc between them and T on the long arc, directly opposite S. Step 1 shows the points and the radii OA and OB. Step 2 highlights the minor arc ASB in blue and shades the 110 degree angle it subtends at O. Step 3 highlights the major arc ATB in orange and shades the reflex angle of 250 degrees at O. Step 4 shows both arcs and notes that 110 degrees and 250 degrees make 360 degrees.

Quick check

1 of 3

The minor arc

A B

of a circle subtends

130^{\circ}

at the centre. What angle does the major arc

A B

subtend at the centre?

Show answer

Answer 230 °

The two angles make a full turn:

360^{\circ} - 130^{\circ} = 230^{\circ}

, a reflex angle.

5.7.1 Angle subtended by an arc at a point on the circle outside the arc

Now pick a point $P$ on the circle but not on the arc $A B$ we are studying. Joining $P$ to $A$ and to $B$ gives $∠ A P B$ , the angle the arc subtends at $P$ . Here is the surprise: slide $P$ anywhere along the rest of the circle and $∠ A P B$ keeps exactly the same size.

ActivityNCERT Activity · p. 107

One arc, many viewpoints

You’ll need

a pair of compasses
a ruler
a protractor

Predict first

Arc

A B

subtends

100^{\circ}

at the centre. What will

∠ A P B

be at a point

P

on the rest of the circle?

100^{\circ}

50^{\circ}

200^{\circ}

it depends on where

P

Draw a circle of radius 5 cm with centre $O$ . With a protractor at $O$ , mark points $A$ and $B$ on the circle so that $∠ A O B = 80^{\circ}$ .
Mark four points $W$ , $X$ , $Y$ and $Z$ , spread out along the major arc.
Join each of them to $A$ and to $B$ , and measure $∠ A W B$ , $∠ A X B$ , $∠ A Y B$ and $∠ A Z B$ .
Repeat everything with $∠ A O B = 140^{\circ}$ .

Table 5.5Angles subtended by one arc

Point	Angle (°) when $∠ A O B = 80^{\circ}$	Angle (°) when $∠ A O B = 140^{\circ}$
$W$
$X$
$Y$
$Z$

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Theorem 9The angle at the centre is double

An arc subtends at the centre an angle twice as large as the angle it subtends at any point on the rest of the circle.

Let $A B$ be the arc, $O$ the centre and $P$ a point on the rest of the circle (Fig. 5.14). Join $P O$ and extend it to meet the circle again at $Q$ . Write $x = ∠ O P A$ and $y = ∠ O P B$ .

Case 1: $Q$ lies on the arc $A B$ . Then $P Q$ runs between $P A$ and $P B$ , so $∠ A P B = x + y$ .

∠ A O B = 2 ∠ A P B

Case 2: $Q$ lies outside the arc $A B$ . Now one of $P A$ , $P B$ runs between the other and $P Q$ ; say $P B$ does, so $∠ A P B = x - y$ . The same two steps still give $∠ A O Q = 2 x$ and $∠ B O Q = 2 y$ , and now $∠ A O B = 2 x - 2 y = 2 (x - y) = 2 ∠ A P B$ .

The argument works whether the arc is minor or major. For a major arc, $∠ A O B$ is the reflex angle and $Q$ always lands on the arc, so Case 1 applies unchanged.

3/4 $∠ A O Q = 2 x$ and $∠ B O Q = 2 y$ , so $∠ A O B = 2 (x + y)$ .

Fig. 5.14Two isosceles triangles and the exterior angle theorem prove Theorem 9.NCERT Figs. 5.21–5.22 · p. 108

A circle with centre O. Step 1: arc AB at the bottom subtends 120 degrees at O and 60 degrees at a point P at the top. Step 2: the diameter from P through O to Q is drawn, and triangles OAP and OBP are shaded, with base angles x, x and y, y because each triangle has two radii as sides. Step 3: the exterior angles at O are marked, angle AOQ equal to 2x and angle BOQ equal to 2y, so angle AOB is 2x plus 2y, twice angle APB. Step 4 shows the second case, with P on the left so that Q falls outside arc AB: now angle AOB is 2x minus 2y and angle APB is x minus y, still exactly half.

Theorem 9 has a striking consequence. Every point $P$ on the rest of the circle sees the arc at the same angle, half of the angle at the centre. The region between a chord $A B$ and one of its arcs is called a segment of the circle (वृत्तखंड), so the fact is usually stated like this.

CorollaryAngles in the same segment

All the angles that an arc subtends at points on the rest of the circle are equal: angles in the same segment are equal.

Step off the circle and the equality breaks (Fig. 5.15). Seen from a point inside the circle (above $A B$ in the figure), the chord looks wider; seen from a point outside, it looks narrower. So the arc is exactly the path along which the view of $A B$ stays the same, an idea we turn around in Section 5.8.

1/3 $P$ , $Q$ and $R$ are on the same arc: each angle is $70^{\circ}$ .

Fig. 5.15Only points on the arc see

A B

at exactly the same angle.NCERT Fig. 5.25 · p. 110

A circle with centre O and a chord AB near the bottom, with the minor arc AB highlighted. Step 1 shows three points P, Q and R on the major arc, each joined to A and B; all three angles are marked 70 degrees. Step 2 fades those points and adds a point I inside the circle, whose angle AIB is about 93 degrees, bigger. Step 3 adds a point E outside the circle, whose angle AEB is about 38 degrees, smaller.

Results like this, which drop out of a theorem almost for free, are called corollaries (उपप्रमेय). The most famous one comes when the arc is a semicircle.

CorollaryAngle in a semicircle

An angle in a semicircle is a right angle: if $A B$ is a diameter and $C$ is any other point of the circle, then $∠ A C B = 90^{\circ}$ .

Take the arc $A B$ that does not contain $C$ . It is a semicircle, and the angle it subtends at the centre is the straight angle $∠ A O B = 180^{\circ}$ (Fig. 5.16). By Theorem 9, $∠ A C B = \frac{1}{2} \times 180^{\circ} = 90^{\circ}$ .

Fig. 5.16The straight angle at the centre halves to a right angle at

C

.NCERT Fig. 5.24 · p. 110

A semicircle on the diameter AB, with the centre O in the middle of AB and the straight angle of 180 degrees shaded at O. A point C on the arc is joined to A and B by blue lines, and a right-angle mark at C shows that angle ACB is 90 degrees.

This also explains the right-angled triangles of Section 5.3. In a right triangle, the hypotenuse subtends $90^{\circ}$ at the opposite vertex, so the arc beyond it subtends $180^{\circ}$ at the centre. The hypotenuse is therefore a diameter of the circumcircle, and the circumcentre is its midpoint.

Example 5.5

One chord, two arcs

Points $A$ and $B$ lie on a circle with centre $O$ , and the minor arc $A B$ subtends $∠ A O B = 124^{\circ}$ . $C$ is a point on the major arc and $D$ is a point on the minor arc. Find $∠ A C B$ and $∠ A D B$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 5.6

A right angle for free

$A B$ is a diameter of a circle and $C$ is a point on the circle with $∠ C A B = 34^{\circ}$ . Find $∠ A C B$ and $∠ C B A$ .

Try it yourself first. Then reveal the solution one step at a time.

Scroll to explore

One arc, many angles

A circle with centre O and points you can drag round it. Depending on the step, the stage shows the angle an arc makes at the centre, the angle it makes at a point P on the circle, two such angles in the same segment, or an angle in a semicircle, with live measurements in a side panel.

Start with the angle at the centre. Drag $B$ round the circle and watch $∠ A O B$ and the length of arc $A B$ change together.

Now a third point $P$ sits on the rest of the circle. Drag $P$ anywhere outside arc $A B$ : $∠ A P B$ does not change, and it is always half of $∠ A O B$ .

Here $P$ has been placed on the short arc between $A$ and $B$ , so it looks at the major arc. The angle at the centre is now reflex, $240^{\circ}$ , and $∠ A P B = 120^{\circ}$ : still exactly half.

Two viewers, $P$ and $Q$ , on the same arc: their angles stay equal however you drag them. Now drag $Q$ across to the other arc. What do the two angles add up to?

Finally, $A$ and $B$ are the ends of a diameter. Wherever $P$ goes, $∠ A P B = 90^{\circ}$ : an angle in a semicircle is a right angle.

Practice · 3 questions

Practice 5.6: questions with answers

1
NumericalEasyNCERT Exercise Set 5.6, Q1
In a circle with centre $O$ , find the length of the chord $P Q$ in each case.
1. $∠ P O Q = 60^{\circ}$ and the radius is 9.5 cm.
  cm
  Show answer
  Answer 9.5 cm
  $O P = O Q$ with $60^{\circ}$ between them, so $△ O P Q$ is equilateral and $P Q = 9.5$ cm.
2. $∠ P O Q = 90^{\circ}$ and the radius is 5 cm.
  cm
  Show answer
  Answer 7.07 cm
  $△ O P Q$ is right-angled at $O$ , so $P Q = \sqrt{5^{2} + 5^{2}} = 5 \sqrt{2} \approx 7.07$ cm.
2
Short answerMediumNCERT Exercise Set 5.6, Q2
$A$ and $B$ are two points on a circle.
1. Suppose $X$ and $Y$ both lie on the circle, on one side of $A B$ . Could $∠ A X B$ and $∠ A Y B$ be different?
  Show solution
  AnswerNo.
  Points of the circle on one side of $A B$ lie on one arc, so $∠ A X B$ and $∠ A Y B$ are angles in the same segment. Each is half the same angle at the centre, so they are equal.
2. Points $X$ and $Y$ of the circle give $∠ A X B = ∠ A Y B$ . Do they have to lie on one side of $A B$ ?
  Show solution
  AnswerNot necessarily.
  If $A B$ is a diameter, every point of the circle, on either side, sees $A B$ at $90^{\circ}$ . For any other chord the angles on the two sides add up to $180^{\circ}$ and are unequal, so for such a chord $X$ and $Y$ must share a side.
3. Now $∠ A X B = ∠ A Y B$ with $X$ and $Y$ on one side of $A B$ , but nobody says $Y$ is on the circle through $A$ , $B$ and $X$ . Does that circle pass through $Y$ ?
  Show solution
  AnswerYes.
  This is exactly Theorem 10 in Section 5.8: if $A B$ subtends equal angles at two points on the same side of it, the four points are concyclic.
NumericalEasyNCERT Exercise Set 5.6, Q3
$P Q R S$ is a quadrilateral with all four vertices on a circle, and $∠ P S R = 112^{\circ}$ . Find $x = ∠ P Q R$ in degrees.
Fig. 5.18
A circle with four points on it joined in order to form the quadrilateral PQRS: P near the top, Q on the right, R near the bottom and S on the left. The angle at S is marked 112 degrees, and the opposite angle at Q is marked with the letter x.
°
Show answer
Answer 68 °
$∠ P S R$ is half the angle that arc $P Q R$ subtends at the centre, and $x$ is half the angle for arc $P S R$ . Those two angles at the centre add up to $360^{\circ}$ , so $112^{\circ} + x = 180^{\circ}$ and $x = 68^{\circ}$ .

5.8 Concyclicity of Points

Three points that are not collinear always lie on one circle (Theorem 1). Four points usually do not: the fourth has to land exactly on the circle through the other three. Points that do lie on one circle are called concyclic (एकवृत्तीय). How can we test four points without drawing any circle?

The same-segment property gives a test. If $C$ and $D$ lie on the circle through $A$ and $B$ , on the same side of $A B$ , then $∠ A C B = ∠ A D B$ . Remarkably, the converse is true as well.

Theorem 10Equal angles on one side mean concyclic

If segment $A B$ subtends equal angles at two points $C$ and $D$ on the same side of line $A B$ , then $A$ , $B$ , $C$ and $D$ lie on one circle.

Theorem 1 gives us a circle through $A$ , $B$ and $C$ . We show that $D$ must lie on it as well, by ruling out every other possibility (Fig. 5.19).

Fig. 5.19If

D

were off the circle, one of these two pictures would apply, and each leads to a contradiction.NCERT Fig. 5.27 · p. 111

Two panels, each showing a circle with a chord AB standing upright on its left side and a point C on the right-hand arc, where angle ACB is marked. In the left panel, labelled Case 1, D lies outside the circle to the lower right; segment AD crosses the circle at E, and the angle at E, marked with a dashed arc, equals the angle at C and is clearly bigger than the angle at D. In the right panel, labelled Case 2, D lies inside the circle and AD extended meets the circle at E; now the angle at D is bigger than the angle at E, which again equals the angle at C.

Suppose $D$ is outside the circle, and segment $A D$ meets the circle again at a point $E$ between $A$ and $D$ , as in the left picture. Then $E$ and $C$ are on the same arc, so $∠ A E B = ∠ A C B$ . But $∠ A E B$ is an exterior angle of $△ E B D$ , so it is bigger than the interior opposite angle $∠ E D B$ , which is $∠ A D B$ . That makes $∠ A C B > ∠ A D B$ , contradicting what we were given. (If $D$ sits where $A D$ misses the circle, a slightly different picture leads to the same contradiction.)

Suppose $D$ is inside the circle. Extend $A D$ beyond $D$ to meet the circle at $E$ . Now $∠ A D B$ is an exterior angle of $△ D B E$ , so $∠ A D B > ∠ A E B = ∠ A C B$ : again a contradiction.

Neither case can happen, so $D$ lies on the circle through $A$ , $B$ and $C$ , and the four points are concyclic.

A quadrilateral whose four vertices are concyclic is called a cyclic quadrilateral (चक्रीय चतुर्भुज). Its angles obey a beautiful rule (Fig. 5.20).

4/4 $140^{\circ} + 220^{\circ} = 360^{\circ}$ , so $∠ A + ∠ C = 180^{\circ}$ .

Fig. 5.20The two arcs make a full turn at the centre, so the opposite angles make half a turn.NCERT Fig. 5.28 · p. 112

A quadrilateral ABCD inscribed in a circle with centre O; angle A is 70 degrees and angle C is 110 degrees. Step 1 shows the quadrilateral and these two angles. Step 2 highlights arc BCD in blue and shades the 140 degree angle it subtends at O; angle A is half of it. Step 3 highlights arc DAB in pink and shades the reflex 220 degree angle it subtends at O; angle C is half of it. Step 4 shows both shaded angles filling the whole turn at O: 140 plus 220 is 360, so angle A plus angle C is 180.

Theorem 11Opposite angles of a cyclic quadrilateral

In a cyclic quadrilateral, each pair of opposite angles adds up to $180^{\circ}$ .

Let $A B C D$ be cyclic, with centre $O$ . $A$ is not on the arc $B C D$ , and $C$ is not on the arc $D A B$ . These two arcs together make up the whole circle.

∠ A + ∠ C = 180^{\circ}

The same argument with the arcs $A B C$ and $C D A$ gives $∠ B + ∠ D = 180^{\circ}$ .

SimulationCyclic quadrilateral

A quadrilateral ABCD whose four vertices you can drag along a circle with centre O, always staying in order. The four angles are marked, and a panel shows each angle together with the sums angle A plus angle C and angle B plus angle D, which stay at 180 degrees however the shape changes.

Fig. 5.21Drag any vertex along the circle and watch the two sums.

Try this

Drag $A$ slowly along the circle. Which angles change, and which sums stay fixed?
Make $A B C D$ look like a rectangle. What must each of its angles be?
Squeeze $B$ very close to $A$ . Is $∠ B + ∠ D$ still $180^{\circ}$ ?

Activity

Tear and fit

You’ll need

a sheet of paper and scissors
a pair of compasses
a ruler

Predict first

You tear off angles

A

and

C

of a cyclic quadrilateral and place them side by side. What do they make together?

a right anglea straight anglea full turn

Draw a large circle and mark any four points on it. Label them $A$ , $B$ , $C$ , $D$ in order round the circle, join them to make a quadrilateral and cut it out.
Tear off the corners at $A$ and $C$ and place them side by side, with their vertices touching.
Do the same with the corners at $B$ and $D$ .
Compare with a friend whose quadrilateral has a different shape.

Can’t do it now? Try the simulation · Fig. 5.21

Example 5.7

Opposite angles in a ratio

In a cyclic quadrilateral $A B C D$ , $∠ A B C$ is three times $∠ A D C$ . Find both angles.

Try it yourself first. Then reveal the solution one step at a time.

The converse of Theorem 11 is also true, and it gives a quick test for four concyclic points.

Theorem 12A test for cyclic quadrilaterals

If one pair of opposite angles of a quadrilateral adds up to $180^{\circ}$ , the quadrilateral is cyclic: one circle passes through all four of its vertices.

Let $A B C D$ be a quadrilateral with $∠ A + ∠ C = 180^{\circ}$ . The four angles add up to $360^{\circ}$ , so $∠ B + ∠ D = 180^{\circ}$ as well; no angle is reflex, and $A$ and $C$ lie on opposite sides of the diagonal $B D$ .

Draw the circle through $A$ , $B$ and $D$ (Theorem 1). We do not yet know whether $C$ is on it (Fig. 5.22).

Fig. 5.22

∠ B E D = 180^{\circ} - ∠ A = ∠ B C D

, so Theorem 10 puts

C

on the circle through

B

D

and

E

.NCERT Fig. 5.29 · p. 113

A circle with centre O drawn through A, B and D. The quadrilateral ABCD is shaded blue, with its fourth vertex C on the right and a question mark beside it. A point E on the arc BD away from A is joined to B and D by dashed lines. The angles at A, C and E are shaded, and a key lists angle A = 75 degrees, angle C = 105 degrees and angle BED = 105 degrees. A note reads: E is on the circle; is C on it too?

Choose any point $E$ on the arc $B D$ that does not contain $A$ . Then $A B E D$ is a cyclic quadrilateral, so Theorem 11 applies to it.

∠ B E D = ∠ B C D

$E$ and $C$ both lie on the side of $B D$ away from $A$ , and $B D$ subtends equal angles at them. By Theorem 10, $B$ , $D$ , $E$ and $C$ lie on one circle. But only one circle passes through $B$ , $D$ and $E$ : the circle we drew. So $C$ is on it, and $A B C D$ is cyclic.

Example 5.8

Is it cyclic?

In quadrilateral $P Q R S$ , $∠ P = 72^{\circ}$ . Side $Q R$ is extended beyond $R$ to a point $T$ , and the exterior angle $∠ S R T = 72^{\circ}$ . Is $P Q R S$ cyclic?

Try it yourself first. Then reveal the solution one step at a time.

That completes our tour. Starting from one simple rule, equal distance from a centre, we have reached twelve theorems about chords, arcs, angles and concyclic points. In Class 10 you will add tangents, lines that touch a circle at exactly one point, and find that they follow rules just as neat as these.

Recap

I’m Up and Down, and Round and Round: key points and summary

Every point of a circle, and no other point of the plane, is at distance $r$ from the centre: a circle is a locus (Section 5.1).
A circle looks the same after any turn about its centre, and every diameter is a line of symmetry (Section 5.2).
Countless circles pass through two given points, and their centres make up the whole perpendicular bisector of those points (Section 5.3).
Exactly one circle, the circumcircle, passes through three non-collinear points; its centre is where the perpendicular bisectors meet (Theorem 1).
Chords of equal length make equal angles at the centre, and chords making equal angles at the centre have equal length (Theorem 2, Theorem 3).
Joining the centre to the middle of a chord gives a right angle, and a perpendicular dropped from the centre cuts a chord in half (Theorem 4, Theorem 5).
Equal chords are equally far from the centre, and chords equally far from the centre are equal (Theorem 6, Theorem 7).
Of two unequal chords the longer is nearer the centre, so a diameter is the longest chord (Theorem 8).
A chord at distance $d$ from the centre has length $2 \sqrt{r^{2} - d^{2}}$ (Example 5.4).
An arc subtends twice as large an angle at the centre as at any point on the rest of the circle (Theorem 9).
Angles in the same segment are equal, and an angle in a semicircle is $90^{\circ}$ (Section 5.7.1).
If $A B$ makes the same angle at $C$ and at $D$ , both on one side of $A B$ , then $A$ , $B$ , $C$ and $D$ are concyclic (Theorem 10).
Opposite angles of a cyclic quadrilateral add up to $180^{\circ}$ , and a quadrilateral with such a pair is cyclic (Theorem 11, Theorem 12).

Flash cards · 13

Theorem flashcards: if … then …

1 / 13

End-of-chapter exercises

Practice · 30 questions

Practice questions with answers (NCERT “End-of-chapter exercises”)

NumericalEasyNCERT End-of-chapter Exercises, Q1
A chord of a circle of radius 17 cm is 8 cm from the centre. How long is the chord?
cm
Show answer
Answer 30 cm
Half-chord $= \sqrt{17^{2} - 8^{2}} = \sqrt{225} = 15$ cm, so the chord is $30$ cm long.
NumericalEasyNCERT End-of-chapter Exercises, Q2
An arc subtends $154^{\circ}$ at the centre of a circle. What angle, in degrees, does it subtend at a point on the rest of the circle?
°
Show answer
Answer 77 °
By Theorem 9 the angle is half: $154^{\circ} \div 2 = 77^{\circ}$ .
NumericalEasyNCERT End-of-chapter Exercises, Q3
A circle has diameter 50 cm. How far from its centre is a chord of length 48 cm?
cm
Show answer
Answer 7 cm
$r = 25$ cm and half the chord is 24 cm, so the distance is $\sqrt{25^{2} - 24^{2}} = \sqrt{49} = 7$ cm.
NumericalEasyNCERT End-of-chapter Exercises, Q4
A circle has radius 29 cm, and one of its chords is 20 cm from the centre. Find the length of the chord.
cm
Show answer
Answer 42 cm
Half the chord is $\sqrt{29^{2} - 20^{2}} = \sqrt{441} = 21$ cm, so the chord is $42$ cm long.
ProofMediumNCERT End-of-chapter Exercises, Q5
A carpenter wants to find the centre of a round table top. Prove that the perpendicular bisector of any chord passes through the centre, and explain how two chords then locate the centre.
Show solution
Let $A B$ be a chord and $O$ the centre. $O A = O B$ (radii), so $O$ is equally far from $A$ and $B$ and therefore lies on the perpendicular bisector of $A B$ (the locus result of Section 5.3).
Draw two chords that are not parallel and construct their perpendicular bisectors. Both pass through $O$ , and two lines that are not parallel meet at only one point, so their crossing point is the centre.
6
Short answerEasyNCERT End-of-chapter Exercises, Q6
$P Q$ is a diameter of a circle and $R$ is a point on the circle with $∠ R P Q = 27^{\circ}$ .
1. Find $∠ P R Q$ in degrees.
  °
  Show answer
  Answer 90 °
  $∠ P R Q$ is an angle in a semicircle, so it is $90^{\circ}$ .
2. Find $∠ R Q P$ in degrees.
  °
  Show answer
  Answer 63 °
  Angle sum in $△ P Q R$ : $180^{\circ} - 90^{\circ} - 27^{\circ} = 63^{\circ}$ .
7
NumericalEasyNCERT End-of-chapter Exercises, Q7
$K L M N$ is a cyclic quadrilateral.
1. If $∠ K = 58^{\circ}$ , find $∠ M$ in degrees.
  °
  Show answer
  Answer 122 °
  Opposite angles add to $180^{\circ}$ : $180^{\circ} - 58^{\circ} = 122^{\circ}$ .
2. If $∠ L = 96^{\circ}$ , find $∠ N$ in degrees.
  °
  Show answer
  Answer 84 °
  $180^{\circ} - 96^{\circ} = 84^{\circ}$ .
8
NumericalMediumNCERT End-of-chapter Exercises, Q8
Quadrilateral $W X Y Z$ is inscribed in a circle, with $∠ W = (3 y + 5)^{\circ}$ and $∠ Y = (2 y + 25)^{\circ}$ .
1. Find $y$ .
  Show answer
  Answer 30
  $∠ W + ∠ Y = 180^{\circ}$ , so $5 y + 30 = 180$ and $y = 30$ .
2. Find $∠ W$ in degrees.
  °
  Show answer
  Answer 95 °
  $3 \times 30 + 5 = 95$ , so $∠ W = 95^{\circ}$ .
3. Find $∠ Y$ in degrees.
  °
  Show answer
  Answer 85 °
  $2 \times 30 + 25 = 85$ , so $∠ Y = 85^{\circ}$ . Check: $95^{\circ} + 85^{\circ} = 180^{\circ}$ .
NumericalMediumNCERT End-of-chapter Exercises, Q9
A chord of length 24 cm is 3.5 cm from the centre of a circle. Find the radius.
cm
Show answer
Answer 12.5 cm
$r = \sqrt{12^{2} + {3.5}^{2}} = \sqrt{144 + 12.25} = \sqrt{156.25} = 12.5$ cm.
NumericalHardNCERT End-of-chapter Exercises, Q10
A cyclic quadrilateral $A B C D$ has $A B = A D = 8$ units and $C B = C D = 15$ units. Find its area in square units.
sq. units
Show solution
Answer120 square units
Triangles $A B C$ and $A D C$ are congruent (SSS: $A B = A D$ , $C B = C D$ and $A C$ is common), so $∠ B = ∠ D$ .
$A B C D$ is cyclic, so $∠ B + ∠ D = 180^{\circ}$ , which gives $∠ B = ∠ D = 90^{\circ}$ .
The quadrilateral is made of two right triangles with legs 8 and 15, so its area is $2 \times \frac{1}{2} \times 8 \times 15 = 120$ square units. (If the sides came in the order 8, 15, 8, 15, the shape would be an 8 by 15 rectangle, with the same area.)
Short answerHardNCERT End-of-chapter Exercises, Q11
Without drawing the circle, how can you tell whether the centre of the circle through the vertices of a cyclic quadrilateral lies inside the quadrilateral, on one of its sides, or outside it?
Show solution
AnswerMeasure the angle each side subtends at a far vertex: the centre is inside exactly when all four of these angles are acute.
Take side $A B$ of the cyclic quadrilateral $A B C D$ . It subtends $∠ A C B$ at the vertex $C$ .
If $∠ A C B < 90^{\circ}$ , the arc $A B$ away from $C$ is a minor arc, so the centre lies on the same side of $A B$ as the quadrilateral. If $∠ A C B = 90^{\circ}$ , $A B$ is a diameter and the centre lies on it. If $∠ A C B > 90^{\circ}$ , the centre lies beyond $A B$ , outside.
So measure $∠ A C B$ , $∠ B D C$ , $∠ C A D$ and $∠ D B A$ , each between a diagonal and a side. The centre is inside the quadrilateral exactly when all four are acute.
ProofHardNCERT End-of-chapter Exercises, Q12
Two equal chords of a circle cross each other at a point $X$ . Show that the two pieces into which $X$ divides one chord are equal to the two pieces of the other chord.
Show solution
Let the chords be $P Q$ and $R S$ , each of length $2 h$ , in a circle with centre $O$ . Drop $O M ⊥ P Q$ and $O N ⊥ R S$ . By Theorems 5 and 6, $M$ and $N$ are midpoints and $O M = O N$ .
Triangles $O M X$ and $O N X$ are right-angled at $M$ and $N$ , share the hypotenuse $O X$ , and have $O M = O N$ . By RHS they are congruent, so $M X = N X$ ; call this length $k$ .
$X$ is $k$ away from the midpoint of each chord, so the pieces of $P Q$ are $h + k$ and $h - k$ , and the pieces of $R S$ are also $h + k$ and $h - k$ .
13
DiagramMediumNCERT End-of-chapter Exercises, Q13
Construct a circle in which a chord of length 8 cm is exactly 3 cm from the centre.
Show solution
AnswerThe circle has radius 5 cm.
Draw $A B = 8$ cm and construct its perpendicular bisector, meeting $A B$ at $M$ .
Mark $O$ on the bisector with $O M = 3$ cm.
Draw the circle with centre $O$ and radius $O A$ .
By construction $O M ⊥ A B$ and $O M = 3$ cm, so $A B$ is a chord 3 cm from the centre. The radius is $O A = \sqrt{4^{2} + 3^{2}} = 5$ cm.
ProofMediumNCERT End-of-chapter Exercises, Q14
A parallelogram has all four vertices on a circle. Prove that it must be a rectangle.
Show solution
In parallelogram $A B C D$ , $∠ A = ∠ C$ . Because $A B C D$ is cyclic, $∠ A + ∠ C = 180^{\circ}$ (Theorem 11).
So $2 ∠ A = 180^{\circ}$ and $∠ A = 90^{\circ}$ . A parallelogram with one right angle has four right angles, so it is a rectangle.
ProofMediumNCERT End-of-chapter Exercises, Q15
A rectangle $A B C D$ is inscribed in a circle. Show that both diagonals are diameters, and hence that they cross at the centre.
Show solution
$∠ A B C = 90^{\circ}$ , so chord $A C$ subtends $90^{\circ}$ at the point $B$ of the circle. By Theorem 9, the arc $A C$ not containing $B$ subtends $2 \times 90^{\circ} = 180^{\circ}$ at the centre $O$ . So $A$ , $O$ and $C$ lie on one line, and $A C$ is a diameter.
In the same way $B D$ is a diameter. Two diameters meet at the centre, so the diagonals cross at $O$ .
Short answerHardNCERT End-of-chapter Exercises, Q16
Consider all the chords of length 16 cm in a circle of radius 10 cm. What figure do their midpoints form?
Show solution
AnswerA circle with the same centre and radius 6 cm.
The midpoint $M$ of each such chord is the foot of the perpendicular from $O$ , so $O M = \sqrt{10^{2} - 8^{2}} = 6$ cm for every one of them.
Conversely, every point 6 cm from $O$ is the midpoint of such a chord: draw the chord through it perpendicular to the radius. So the midpoints form the circle with centre $O$ and radius 6 cm.
ProofMediumNCERT End-of-chapter Exercises, Q17
Chords $P Q$ and $P R$ of a circle with centre $O$ are equal. Prove that $O$ lies on the bisector of $∠ Q P R$ .
Show solution
$O Q = O R$ (radii), $P Q = P R$ (given) and $O P$ is common, so $△ O P Q ≅ △ O P R$ (SSS).
Hence $∠ O P Q = ∠ O P R$ : the line $P O$ bisects $∠ Q P R$ , so $O$ lies on the bisector.
NumericalHardNCERT End-of-chapter Exercises, Q18
Two parallel chords of a circle, 14 cm and 48 cm long, lie on the same side of the centre and are 17 cm apart. Find the radius.
cm
Show solution
The half-chords are 7 cm and 24 cm. The longer chord is nearer the centre (Theorem 8); let its distance be $d$ , so the shorter chord is $d + 17$ from the centre.
$r^{2} = d^{2} + 24^{2} = (d + 17)^{2} + 7^{2}$
Expanding: $d^{2} + 576 = d^{2} + 34 d + 289 + 49$ , so $34 d = 238$ and $d = 7$ .
$r = \sqrt{7^{2} + 24^{2}} = \sqrt{625} = 25 c m$
19
NumericalHardNCERT End-of-chapter Exercises, Q19
The six vertices of a regular hexagon lie on a circle of radius 8 cm.
1. How long is each side?
  cm
  Show answer
  Answer 8 cm
  Each side subtends $360^{\circ} \div 6 = 60^{\circ}$ at the centre, so each of the six triangles formed with the centre is equilateral, and each side is 8 cm.
2. How far is each side from the centre?
  cm
  Show answer
  Answer 6.93 cm
  Distance $= \sqrt{8^{2} - 4^{2}} = \sqrt{48} = 4 \sqrt{3} \approx 6.93$ cm.
20
Short answerMediumNCERT End-of-chapter Exercises, Q20
Quadrilateral $W X Y Z$ is inscribed in a circle, with $W X$ a diameter and $∠ X W Z = 38^{\circ}$ . Give a reason for each answer.
1. Find $∠ W Z X$ in degrees.
  °
  Show answer
  Answer 90 °
  $∠ W Z X$ is subtended by the diameter $W X$ at the point $Z$ of the circle: an angle in a semicircle, so it is $90^{\circ}$ .
2. Find $∠ W X Z$ in degrees.
  °
  Show answer
  Answer 52 °
  Angle sum in $△ W X Z$ : $180^{\circ} - 90^{\circ} - 38^{\circ} = 52^{\circ}$ .
3. Find $∠ W Y Z$ in degrees.
  °
  Show solution
  Both $∠ W Y Z$ and $∠ W X Z$ are subtended by the chord $W Z$ , at points $Y$ and $X$ that lie on the same arc $W X Y Z$ . Angles in the same segment are equal, so $∠ W Y Z = 52^{\circ}$ .
4. Find $∠ W Y X$ in degrees.
  °
  Show answer
  Answer 90 °
  $∠ W Y X$ is also subtended by the diameter $W X$ , so it is an angle in a semicircle: $90^{\circ}$ .
ProofMediumNCERT End-of-chapter Exercises, Q21
$P Q R S$ is a cyclic quadrilateral, and side $Q R$ is extended beyond $R$ to a point $T$ . Prove that the exterior angle $∠ S R T$ equals the interior opposite angle $∠ Q P S$ .
Show solution
$∠ S R T = 180^{\circ} - ∠ Q R S$ (angles on a straight line).
$∠ Q P S = 180^{\circ} - ∠ Q R S$ (Theorem 11, since $P$ and $R$ are opposite vertices).
So $∠ S R T = ∠ Q P S$ .
ProofMediumNCERT End-of-chapter Exercises, Q22
Justify the statement: no chord of a circle is longer than a diameter.
Show solution
Let $A B$ be a chord and $O$ the centre. If $O$ is not on $A B$ , the triangle inequality in $△ O A B$ gives $A B < O A + O B = 2 r$ .
If $O$ is on $A B$ , then $A B = O A + O B = 2 r$ , and the chord is a diameter. So every chord is at most $2 r$ long. (Equivalently, $2 \sqrt{r^{2} - d^{2}} \leq 2 r$ , with equality only when $d = 0$ .)
ProofHardNCERT End-of-chapter Exercises, Q23
$K$ is a point inside a circle with centre $O$ , and $K \neq O$ . Show that the shortest chord through $K$ is the one perpendicular to $O K$ .
Show solution
Let a chord through $K$ be at distance $d$ from $O$ , with $M$ the foot of the perpendicular. If $M \neq K$ , triangle $O M K$ has its right angle at $M$ , so the hypotenuse $O K$ is longer than $O M$ . So $d \leq O K$ , with equality exactly when $M = K$ , that is, when the chord is perpendicular to $O K$ .
The chord's length is $2 \sqrt{r^{2} - d^{2}}$ , which is smallest when $d$ is largest. So the shortest chord through $K$ is the one at distance $O K$ from the centre: the chord perpendicular to $O K$ .
ProofMediumNCERT End-of-chapter Exercises, Q24
$A B$ is a diameter of a circle with centre $O$ , and $C$ is another point on the circle. Join $O C$ , and call the base angles of the isosceles triangles $O A C$ and $O B C$ $p$ and $q$ . Use them to show that $∠ A C B = 90^{\circ}$ without using Theorem 9.
Show solution
$O A = O C$ , so $∠ O A C = ∠ O C A = p$ ; and $O B = O C$ , so $∠ O B C = ∠ O C B = q$ .
$∠ A C B = ∠ O C A + ∠ O C B = p + q$ , so the angles of $△ A B C$ are $p$ , $q$ and $p + q$ , which add up to $2 (p + q) = 180^{\circ}$ .
Hence $∠ A C B = p + q = 90^{\circ}$ .
ProofHardNCERT End-of-chapter Exercises, Q25
$A B$ is a diameter of a circle. Chords $P P^{'}$ and $Q Q^{'}$ are both perpendicular to $A B$ . Prove that the segment joining the midpoint of $P Q$ to the midpoint of $P^{'} Q^{'}$ is perpendicular to $A B$ .
Show solution
Reflection in the diameter $A B$ maps the circle onto itself (Section 5.2). $A B$ passes through the centre and is perpendicular to $P P^{'}$ , so it bisects $P P^{'}$ (Theorem 5); the reflection therefore swaps $P$ and $P^{'}$ . In the same way it swaps $Q$ and $Q^{'}$ .
So the reflection carries the segment $P Q$ onto $P^{'} Q^{'}$ , and the midpoint $M$ of $P Q$ onto the midpoint $M^{'}$ of $P^{'} Q^{'}$ .
A point and its mirror image are joined by a segment perpendicular to the mirror line, so $M M^{'} ⊥ A B$ . (If $M$ happens to lie on $A B$ , then $M^{'} = M$ and there is nothing to prove.)
ProofMediumNCERT End-of-chapter Exercises, Q26
$A B C D$ is a cyclic quadrilateral whose centre $O$ lies inside it. Join $O$ to the four vertices, and call the base angles of the isosceles triangles $O A B$ , $O B C$ , $O C D$ and $O D A$ $w$ , $x$ , $y$ and $z$ . Use them to prove that $∠ A + ∠ C = 180^{\circ}$ .
Show solution
Each vertex angle splits into two base angles: $∠ A = z + w$ , $∠ B = w + x$ , $∠ C = x + y$ and $∠ D = y + z$ .
The angles of the quadrilateral add up to $360^{\circ}$ , so $2 (w + x + y + z) = 360^{\circ}$ and $w + x + y + z = 180^{\circ}$ .
Hence $∠ A + ∠ C = z + w + x + y = 180^{\circ}$ .
Quick answerEasy
Which statement is false?
Equal chords of a circle are equally far from its centre.A perpendicular dropped from the centre onto a chord cuts it in half.An angle in a semicircle is a right angle.A longer chord is farther from the centre.
Show answer
Answer (D) A longer chord is farther from the centre.
The longer chord is nearer the centre (Theorem 8). The other three are Theorem 6, Theorem 5 and the semicircle corollary.
Quick answerEasy
Arc $P Q$ subtends $∠ P R Q = 41^{\circ}$ at a point $R$ on the rest of the circle. The angle it subtends at the centre is degrees.
Show answer
Answer 82
Twice $41^{\circ}$ is $82^{\circ}$ .
29
Quick answerMedium
Put these chords of a circle of radius 5 cm in order, from nearest to the centre to farthest from it.
1. the 8 cm chord
2. the 6 cm chord
3. the 10 cm chord
4. the 2.8 cm chord
Show answer
Answer
the 10 cm chord
the 8 cm chord
the 6 cm chord
the 2.8 cm chord
Their distances are 0, 3, 4 and 4.8 cm: the longer the chord, the nearer it is.
30
Quick answerMedium
Match each theorem with the congruence rule used in its proof in this chapter.
Tap an item on the left, then its match on the right.
Show answer
Answer
Theorem 2: equal chords, equal angles : SSS
Theorem 3: equal angles, equal chords : SAS
Theorem 5: the perpendicular bisects the chord : RHS
Theorem 2 gives three pairs of sides, Theorem 3 two sides and the angle between them, and Theorem 5 a right angle, a hypotenuse and a side.

Find your NCERT exercise question

Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
NCERT Exercise Set 5.1, Q1	Practice 5.1 · Q1
NCERT Exercise Set 5.1, Q2	Practice 5.1 · Q2
NCERT Exercise Set 5.1, Q3	Practice 5.1 · Q3
NCERT Exercise Set 5.1, Q4	Practice 5.1 · Q4
NCERT Ex. Set 5.1, Think, Draw and Infer Q1	Practice 5.1 · Q5
NCERT Ex. Set 5.1, Think, Draw and Infer Q2	Practice 5.1 · Q6
NCERT Exercise Set 5.2, Q1	Practice 5.2 · Q1
NCERT Exercise Set 5.2, Q2	Practice 5.2 · Q2
NCERT Exercise Set 5.3, Q1	Practice 5.3 · Q1
NCERT Exercise Set 5.3, Q2	Practice 5.3 · Q2
NCERT Exercise Set 5.3, Q3	Practice 5.3 · Q3
NCERT Exercise Set 5.4, Q1	Practice 5.4 · Q1
NCERT Exercise Set 5.4, Q2	Practice 5.4 · Q2
NCERT Exercise Set 5.4, Q3	Practice 5.4 · Q3
NCERT Exercise Set 5.5, Q1	Practice 5.5 · Q1
NCERT Exercise Set 5.5, Q2	Practice 5.5 · Q2
NCERT Exercise Set 5.5, Q3	Practice 5.5 · Q3
NCERT Exercise Set 5.6, Q1	Practice 5.6 · Q1
NCERT Exercise Set 5.6, Q2	Practice 5.6 · Q2
NCERT Exercise Set 5.6, Q3	Practice 5.6 · Q3
NCERT End-of-chapter Exercises, Q1	End-of-chapter exercises · Q1
NCERT End-of-chapter Exercises, Q2	End-of-chapter exercises · Q2
NCERT End-of-chapter Exercises, Q3	End-of-chapter exercises · Q3
NCERT End-of-chapter Exercises, Q4	End-of-chapter exercises · Q4
NCERT End-of-chapter Exercises, Q5	End-of-chapter exercises · Q5
NCERT End-of-chapter Exercises, Q6	End-of-chapter exercises · Q6
NCERT End-of-chapter Exercises, Q7	End-of-chapter exercises · Q7
NCERT End-of-chapter Exercises, Q8	End-of-chapter exercises · Q8
NCERT End-of-chapter Exercises, Q9	End-of-chapter exercises · Q9
NCERT End-of-chapter Exercises, Q10	End-of-chapter exercises · Q10
NCERT End-of-chapter Exercises, Q11	End-of-chapter exercises · Q11
NCERT End-of-chapter Exercises, Q12	End-of-chapter exercises · Q12
NCERT End-of-chapter Exercises, Q13	End-of-chapter exercises · Q13
NCERT End-of-chapter Exercises, Q14	End-of-chapter exercises · Q14
NCERT End-of-chapter Exercises, Q15	End-of-chapter exercises · Q15
NCERT End-of-chapter Exercises, Q16	End-of-chapter exercises · Q16
NCERT End-of-chapter Exercises, Q17	End-of-chapter exercises · Q17
NCERT End-of-chapter Exercises, Q18	End-of-chapter exercises · Q18
NCERT End-of-chapter Exercises, Q19	End-of-chapter exercises · Q19
NCERT End-of-chapter Exercises, Q20	End-of-chapter exercises · Q20
NCERT End-of-chapter Exercises, Q21	End-of-chapter exercises · Q21
NCERT End-of-chapter Exercises, Q22	End-of-chapter exercises · Q22
NCERT End-of-chapter Exercises, Q23	End-of-chapter exercises · Q23
NCERT End-of-chapter Exercises, Q24	End-of-chapter exercises · Q24
NCERT End-of-chapter Exercises, Q25	End-of-chapter exercises · Q25
NCERT End-of-chapter Exercises, Q26	End-of-chapter exercises · Q26

Quick check: MCQs on I’m Up and Down, and Round and Round

Warm-up2 questions
5.1 Definitions4 questions
5.2 Symmetries of a Circle4 questions
5.3 How Many Circles?4 questions
5.4 Chords and the Angles They Subtend3 questions
5.5 Midpoints and Perpendicular Bisectors of Chords4 questions
5.6.1 Which of the two unequal chords is farther from the centre?4 questions
5.7 Angles Subtended by an Arc3 questions
5.7.1 Angle subtended by an arc at a point on the circle outside the arc4 questions
5.8 Concyclicity of Points4 questions

Glossary · 15

Key terms (English–Hindi)

arcचापchaap: The part of a circle traced when moving along it from one point to another. The shorter of the two arcs is the minor arc, the longer the major arc.First seen in the chapter →

chordजीवाjeeva: A segment whose two ends lie on the circle.First seen in the chapter →
circleवृत्तvritt: A round curve whose points are all exactly $r$ from one point inside it, the centre; $r$ is the radius.
circumcentreपरिकेंद्रparikendra: The centre of a circumcircle: where the perpendicular bisectors of the sides of a triangle meet.First seen in the chapter →
circumcircleपरिवृत्तparivritt: The one circle that passes through all three vertices of a triangle.
collinearसंरेखsanrekh: Lying on one straight line.First seen in the chapter →
concyclicएकवृत्तीयekvrittiya: Points are concyclic when one circle passes through all of them.First seen in the chapter →
corollaryउपप्रमेयupapramey: A result that follows almost at once from a theorem just proved.First seen in the chapter →
cyclic quadrilateralचक्रीय चतुर्भुजchakriya chaturbhuj: A quadrilateral whose four vertices lie on one circle.First seen in the chapter →

diameterव्यासvyaas: A chord through the centre; its length is $2 r$ and no chord is longer.First seen in the chapter →

locusबिंदुपथbindupath: The set of exactly those points that obey a given rule, and no others.First seen in the chapter →

perpendicular bisectorलंब समद्विभाजकlamb samadvibhajak: The line through the midpoint of a segment at right angles to it; its points are exactly those equally far from the two ends.First seen in the chapter →

radiusत्रिज्याtrijya: A segment from the centre to a point of the circle, or its length $r$ .First seen in the chapter →

segment of a circleवृत्तखंडvrittkhand: The region between a chord and one of the two arcs it cuts off.First seen in the chapter →
subtendअंतरित करनाantarit karna: A chord or arc $A B$ subtends $∠ A P B$ at a point $P$ : the angle formed by joining $P$ to $A$ and to $B$ .First seen in the chapter →

Circles all around us

5.1 Definitions

5.2 Symmetries of a Circle

5.3 How Many Circles?

5.4 Chords and the Angles They Subtend

5.5 Midpoints and Perpendicular Bisectors of Chords

5.6 Distance of Chords from the Centre

5.6.1 Which of the two unequal chords is farther from the centre?

5.7 Angles Subtended by an Arc

5.7.1 Angle subtended by an arc at a point on the circle outside the arc

5.8 Concyclicity of Points

End-of-chapter exercises

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