Lesson: Circle

Exercise 10.1

Question: 1

Fill in the blanks:

(i) The centre of a circle lies in the____________ of

the circle. (exterior/ interior)

(ii) A point, whose distance from the centre of a circle

is greater than its radius lies in the__________ of the

circle. (exterior/ interior)

(iii) The longest chord of a circle is a _____________

of the circle.

(iv) An arc is a ___________ when its ends are the

ends of a diameter.

(v) Segment of a circle is the region between an arc

and a_____________ of the circle.

(vi) A circle divides the plane, on which it lies, in

_____________ parts.

Solution

(i) The centre of a circle lies in the interior of the

circle.

(ii) A point, whose distance from the centre of a circle

is greater than its radius lies in the exterior of the

circle.

(iii) The longest chord of a circle is a diameter of the

circle.

(iv) An arc is a semicircle when its ends are the ends

of a diameter.

(v) Segment of a circle is the region between an arc

and a chord of the circle.

(vi) A circle divides the plane, on which it lies,

in three parts.

Question: 2

Write True or False. Give reasons for your answers.

(i) Line segment joining the centre to any point on the

circle is a radius of the circle.

(ii) A circle has only finite number of equal chords.

(iii) If a circle is divided into three equal arcs, each of

them is a major arc.

(iv) A chord of a circle, which is twice as long as its

radius, is a diameter of the circle.

(v) Sector is the region between the chord and its

corresponding arc.

(vi) A circle is a plane figure.

Solution

(i) True.

All line segments from the centre of the circle to the

point on the circle are of equal length and are called

the radius of the circle.

(ii) False.

There can be infinite numbers of equal chords.

(iii) False.

Let us consider a circle with three equal arcs PQ, QR

and PR as shown in the figure below.

Now, for arc PQ, the major arc is PRQ and the same is

true for the other two arcs.

Hence, all the three arcs are minor arcs.

(iv) True.

A chord which is twice as long as a radius must pass

through the centre of the circle. Hence, it is called the

diameter of the circle.

(v) False.

Sector is the region between the two radii of the circle

and the arc.

(vi) True.

A circle is a two-dimensional figure and can be drawn

on the plane.

Exercise 10.2

Question: 1

Recall that two circles are congruent if they have the

same radii. Prove that equal chords of congruent

circles subtend equal angles at their centres.

Solution

Given two congruent circles.

Let AB and CD be two equal chords of two congruent

circles having centres O and O’ respectively.

AB CD

(Equal chords)

Now,

In ΔAOB and ΔCOD,

’OA OC

(Radii)

OB OD

(Radii)

AB CD

(Given)

Δ ΔAOB COD

(SSS congruence condition)

Thus,

AOB COD

by CPCT.

Hence, equal chords of congruent circles subtend

equal angles at their centres.

Question: 2

Prove that if chords of congruent circles subtend equal

angles at their centres, then the chords are equal.

Solution

Given two congruent circles with equal radius and

centres O and

O

.

Given,

AOB COD

(Equal angles)

Now, in ΔAOB and

ΔCOD

,

OA OC

(Radii)

AOB COD

(Given)

OB OD

(Radii)

Δ ΔAOB COD

(SAS congruence condition)

Thus,

AB CD

by CPCT.

Hence, if chords of congruent circles subtend equal

angles at their centres, then the chords are equal.

Exercise 10.3

Question: 1

Draw different pairs of circles. How many points does

each pair have in common? What is the maximum

number of common points?

Solution

No point in common.

No point in common.

One point P is common.

One point P is common.

Two points, P and Q, are common.

Question: 2

Suppose you are given a circle. Give a construction to

find its centre.

Solution

Construction of a circle to find its centre:

Step I: Draw a circle.

Step II: Draw two chords AB and CD as shown in the

figure.

Step III: Draw perpendicular bisector of the chords AB

and CD.

Step IV: Let these two perpendicular bisectors meet at

a point ‘O’. Hence, O is the centre of the given circle.

Question: 3

If two circles intersect at two points, prove that their

centres lie on the perpendicular bisector of the

common chord.

Solution

Let two circles with centres as O and

O

intersect

each other at points P and Q.

We need to prove that

OO

is the perpendicular

bisector of PQ.

Now, in

ΔPOO

and

Δ 'QOO

,

OP OQ

(Radii)

' 'OO OO

(Common)

' 'O P O Q

(Radii)

Δ ' Δ 'POO QOO

(SSS congruence condition)

Hence,

' 'POO QOO

--- (i)

Again, in

ΔPOR

and

ΔQOR

,

OP OQ

(Radii)

POR QOR

(From equation (i))

OR OR

(Common)

Δ ΔPOR QOR

(SAS congruence condition)

Hence,

PRO QRO

and PR = RQ (By CPCT)

also,

180PRO QRO

180

90

2

PRO QRO

Therefore,

OO' is the perpendicular bisector of PQ.

Exercise 10.4

Question: 1

Two circles of radii 5 cm and 3 cm intersect at two

points and the distance between their centres is 4 cm.

Find the length of the common chord.

Solution

Let two circles with centres as ‘O’ and ‘S’ intersect at

points P and Q as shown in the figure.

Given,

5 cm, 3 cmOP PS

and

4 cmOS

.

also,

2PQ PR

Let

RS

be a.

Therefore, if two circles intersect at two points, then

their centres lie on the perpendicular bisector of the

common chord.

90ORP PRS

Now, in

ΔPOR

,

2 2 2

OP OR PR

(By Pythagoras theorem)

2

2 2

2 2

2 2

5 4

25 16 8

9 8 i

a PR

a a PR

PR a a

In

ΔPRS

,

2 2 2

2 2 2

2 2

3

9 ... ii

PS PR RS

PR a

PR a

Equating (i) and (ii),

2 2

9 8 9

8 0

0

a a a

a

a

Putting the value of a in (ii), we get,

2 2

9 0

3 cm

PR

PR

Length of the chord

2 2 3 6 cmPQ PR

Question: 2

If two equal chords of a circle intersect within the

circle, prove that the segments of one chord are equal

to the corresponding segments of the other chord.

Solution

Let AB and CD be the chords of a circle with centre O,

intersecting at point E as shown in the figure.

Also,

AB CD

(Given)

Now, draw a perpendicular line from the centre of the

circle to chords AB and CD, such that,

OM ⊥ AB and ON ⊥ CD.

Also, join O to point E.

Now,

OM bisects

( )AB OM AB

ON bisects

( )CD ON CD

And

AB CD

, therefore,

... iAM ND

and

... iiMB CN

In

ΔOME

and

ΔONE

,

OME ONE

(Perpendiculars)

OE OE

(Common)

OM ON

(

AB CD

and therefore equidistant from

the centre)

Δ ΔOME ONE

by RHS congruence condition.

ME EN

by CPCT --- (iii)

From (i) and (iii), we get,

AM ME ND EN

AE ED

From (ii) and (iii), we get,

MB ME CN EN

EB CE

Hence, it is proved.

Question: 3

If two equal chords of a circle intersect within the

circle, prove that the line joining the point of

intersection to the centre makes equal angles with the

chords.

Solution

Let AB and CD be the chords of a circle, with centre O,

intersecting at point E as shown in the figure.

Also,

AB CD

(Given)

Now, draw a perpendicular line from the centre of the

circle to the chords AB and CD, such that,

OM AB

and

ON CD

.

Also, join O to point E. PQ is the diameter.

Now, in

ΔOEM

and

ΔOEN

,

OM ON

(∵ Equal chords are equidistant from the

centre)

OE OE

(Common)

OME ONE

(Perpendicular)

Δ ΔOEM OEN

by RHS congruence condition.

Therefore,

MEO NEO

(By CPCT)

BEQ CEQ

Question: 4

If a line intersects two concentric circles (circles with

the same centre) with centre O at A, B, C and D, prove

that

AB CD

(See Fig. 10.25).

Solution

Draw a perpendicular OM to line AD from point ‘O’.

OM

bisects

AD

as

OM AD

.

... iAM MD

Also,

OM

bisects

BC

as

OM BC

.

... iiBM MC

From (i) and (ii),

AM BM MD MC

AB CD

Hence, it is proved.

Question: 5

Three girls Reshma, Salma and Mandip are playing a

game by standing on a circle of radius 5 m drawn in a

park. Reshma throws a ball to Salma, Salma to

Mandip, Mandip to Reshma. If the distance between

Reshma and Salma and between Salma and Mandip is

6 m each, what is the distance between Reshma and

Mandip?

Solution

Let Reshma, Salma and Mandip stand at positions A,

B and C on a circle with centre at O.

6 mAB

and

6 mBC

(Given)

Radius

5 mOA

(Given)

Draw a perpendicular BM to AC as shown in the

figure.

∵ ABC is an isosceles triangle, then

AB BC

and M is

the mid-point of AC and BM is a perpendicular

bisector of AC. Hence, BM passes through the centre

of the circle.

Now, let

AM y

and

OM x

then

5BM x

.

2 2 2

OA OM AM

(By Pythagoras theorem)

2 2 2

5 ... ix y

Also,

2 2 2

AB BM AM

(By Pythagoras theorem)

2

2 2

6 5 ... iix y

Subtracting (i) from (ii), we get

2

2

36 25 5

11 25 10

10 14

7

5

x x

x

x

x

Substituting the value of x in (i), we get

2

2

2

2

49

25

25

49

25

25

625 49

25

576

25

24

5

y

y

y

y

y

Therefore,

24 48

2 2 2 m m 9.6 m

5 5

AC AM y

⇒ Distance between Reshma and Mandip is 9.6 m.

Question: 6

A circular park of radius 20 m is situated in a colony.

Three boys Ankur, Syed and David are sitting at equal

distance on its boundary each carrying a toy telephone

in his hand and talking to each other. Find the length

of the string of each phone.

Solution

Let Ankur, Syed and David be sitting at positions A, B,

and C on the boundary of a circular park with centre at

O and radius as 20 m.

All three boys are sitting at equal distances from each

other. Thus, ABC is an equilateral triangle.

Draw a perpendicular AD to BC. Also, AD is the

median of ΔABC and it passes through the centre O.

Again, O is the centroid of ΔABC. OA is the radius of

the circle.

2

3

OA AD

Let the side of a triangle be a metres.

2

a

BD

Applying Pythagoras theorem in ΔABD,

2 2 2

2 2 2

2

2 2

2

2

2

3

3 / 2, 2/ 3

20 2/ 3 3 / 2

20 3

4

AB BD AD

AD AB BD

a

AD a

a

AD

AD a OA AD

m a

a m

Length of the string is 20√3 m.

Exercise 10.5

Question: 1

In Fig. 10.36, A, B and C are three points on a circle

with centre O such that

30BOC

and

60AOB

. If D is a point on the circle other than

the arc ABC, find

ADC

.

Solution

Given,

30BOC

and

60AOB

60 30

90

AOC AOB BOC

AOC

AOC

∵ The angle subtended by an arc at the centre is

double the angle subtended by it at any point on the

remaining part of the circle,

1 1

90 45

2 2

ADC AOC

Question: 2

A chord of a circle is equal to the radius of the circle.

Find the angle subtended by the chord at a point on the

minor arc and also at a point on the major arc.

Solution

Let OA and OB be the radius of the circle with centre

as O, and AB as the chord of a circle.

OA OB AB

radius of the circle (Given)

Therefore,

ΔOAB

is an equilateral triangle.

60AOB

Also,

1 1

60 30

2 2

ACB AOB

ACBD is a cyclic quadrilateral,

180ACB ADB

(Opposite angles of cyclic

quadrilateral are supplementary.)

180 30 150ADB

Hence, angle subtended by the chord at a point on the

minor arc and at a point on the major arc are 150° and

30° respectively.

Question: 3

In Fig. 10.37,

100PQR

, where P, Q and R are

points on a circle with centre O. Find

OPR

.

Solution

Let us take any point S on the major arc of chord PR

as shown in the figure.

Join PS and RS.

∵ Quadrilateral PQRS is a cyclic quadrilateral.

180PQR PSR

(Sum of opposite angles of a

cyclic quadrilateral is supplementary.)

180 100 80PSR

(Given,

100PQR

) …. (i)

Now, the angle subtended by an arc at the centre is

double the angle subtended by it at any point on the

remaining part of the circle.

Here, arc PR is subtending ∠POR at the centre and

∠PSR at point S on the remaining part of the circle.

2 2 80 160POR PSR

(Using equation

(i)) … (ii)

In ∆POR,

OP OR

(Each being the radius)

ORP OPR

(Angles opposite to equal sides

in a triangle) … (iii)

Also,

180POR OPR ORP

(Angle sum

property of a triangle)

160 180OPR OPR

(Using equations (ii)

and (iii))

10OPR

Question: 4

In Fig. 10.38,

69 , 31 ,ABC ACB

find

BDC

.

Solution

In

ΔABC

and

ΔADC

,

BAC BDC

(Angles in the segment of the circle

are equal.)

Now, in

ΔABC

,

180BAC ABC ACB

(Sum of the angles in

a triangle)

69 31 180

180 100

80

BAC

BAC

BAC

Therefore,

80BDC

.

Question: 5

In Fig. 10.39, A, B, C and D are four points on a circle.

AC and BD intersect at a point E such that

130BEC

and

20 .ECD

Find

BAC

.

Fig. 10.39

Solution

In

ΔABC

and

Δ ,ADC

BAC CDE

(Angles in the segment of the circle

are equal.)

In

ΔCDE

,

CEB CDE DCE

(Exterior angles of a triangle

is the sum of the interior opposite angles.)

130 20

110

CDE

CDE

Therefore,

110BAC

.

Question: 6

ABCD is a cyclic quadrilateral whose diagonals

intersect at a point E. If

70 ,DBC BAC

=

30

,

find

BCD

. Further, if

AB BC

, find

ECD

.

Solution

In

ΔADC

and

ΔDBC

,

CBD CAD

(Angles in same segment of a circle

are equal.)

70CAD

30 70 100BAD BAC CAD

180 BCD BAD

(Opposite angles of a cyclic

quadrilateral)

In

100 180

80

Δ

BCD

BCD

ABC

AB BC

(Given)

BCA CAB

(Angles opposite to equal sides of a

triangle)

30BCA

also,

80BCD

80

30 80

50

50

BCA ACD

ACD

ACD

ECD

Question: 7

If diagonals of a cyclic quadrilateral are diameters of

the circle through the vertices of the quadrilateral,

prove that it is a rectangle.

Solution

Let ABCD be a cyclic quadrilateral as shown in figure

below and its diagonals AC and BD be the diameters

of the circle through the vertices of the quadrilateral.

90ABC BCD CDA DAB

(Angles in

the semi-circle)

Therefore, ABCD is a rectangle as each internal angle

is 90°.

Question: 8

If the non-parallel sides of a trapezium are equal,

prove that it is cyclic.

Solution

Let AD and BC be the non-parallel sides of trapezium

ABCD, such that,

AD BC

Let us draw DM and CN perpendicular to AB.

Now, in

ΔDAM

and

Δ ,CBN

AD BC

(Given)

AMD BNC

(Right angles)

DM CN

(Distance between the parallel lines)

Δ ΔDAM CBN

by RHS congruence condition.

Now,

A B

by CPCT

also,

180B C

(Sum of the co-interior angles)

180A C

Therefore, ABCD is a cyclic quadrilateral as sum of

the pair of opposite angles is 180°.

Question: 9

Two circles intersect at two points B and C. Through

B, two line segments ABD and PBQ are drawn to

intersect the circles at A, D and P, Q respectively (see

Fig. 10.40). Prove that

ACP QCD

.

Solution

Let us join A and P, and D and Q as shown in the

figure below.

Now, for chord AP,

PBA ACP

(∵ Angles in the same segment) --- (i)

Similarly, for chord DQ,

DBQ QCD

(Angles in same segment) --- (ii)

ABD and PBQ are line segments intersecting at B.

PBA DBQ

(Vertically opposite angles) --- (iii)

By the equations (i), (ii) and (iii),

ACP QCD

Hence, it is proved.

Question: 10

If circles are drawn taking two sides of a triangle as

diameters, prove that the point of intersection of these

circles lies on the third side.

Solution

Let the triangle be ABC. Now, let us draw two circles

on the sides AB and AC of the

ΔABC

as diameters

such that the circles intersect at point D.

Join AD. Now, to prove that D lies on BC, we need to

prove that BDC is a straight line.

90ADB ADC

(Angle in the semi-circle)

Now,

180ADB ADC

BDC

is straight line.

Therefore, D lies on the side BC.

Question: 11

ABC and ADC are two right triangles with common

hypotenuse AC. Prove that

CAD CBD

.

Solution

Let

ΔABC

and

ΔADC

be the two right angled

triangles such that hypotenuse AC is common.

90ABC ADC

(Given)

Since,

ABC

and

ADC

are 90°, these angles are

in the semi circle. Hence, both the triangles lie on the

semi circle and AC is the diameter of the circle.

⇒ Points A, B, C and D are concyclic.

Thus, CD is the chord.

CAD CBD

(Angles in the same segment of

the circle)

Hence, it is proved.

Question: 12

Prove that a cyclic parallelogram is a rectangle.

Solution

Let ABCD be a cyclic parallelogram as shown in the

figure.

1 2 180

(Opposite angles of a cyclic

parallelogram)

Also, opposite angles of a cyclic parallelogram are

equal.

Therefore,

1 2

1 1 180

2 1 90

One of the interior angle of the parallelogram is right

angled. Therefore, ABCD is a rectangle.