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)

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

_____________ parts.

Solution

circle.

is greater than its radius lies in the exterior of the

of a diameter.

and a chord of the circle.

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.

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.

(iv) True.

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

and the arc.

(vi) True.

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

on the plane.

Exercise 10.2

Question: 1

Solution

Given two congruent circles.

Let AB and CD be two equal chords of two congruent

circles having centres O and O’ respectively.

(SSS congruence condition)

Thus,

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

.

Given,

(Given)

(SAS congruence condition)

Thus,

Hence, if chords of congruent circles subtend equal

angles at their centres, then the chords are equal.

Exercise 10.3

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:

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

Question: 3

If two equal chords of a circle intersect within the

circle, prove that the line joining the point of

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

intersecting at point E as shown in the figure.

Also,

Now, draw a perpendicular line from the centre of the

circle to the chords AB and CD, such that,

and

.

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

Now, in

and

,

(∵ Equal chords are equidistant from the

centre)

(Common)

(Perpendicular)

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

Solution

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

bisects

as

From (i) and (ii),

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

Solution

Let Reshma, Salma and Mandip stand at positions A,

B and C on a circle with centre at O.

∵ ABC is an isosceles triangle, then

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.

(By Pythagoras theorem)

Subtracting (i) from (ii), we get

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

Therefore,

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.

Let the side of a triangle be a metres.

Applying Pythagoras theorem in ΔABD,

. If D is a point on the circle other than

the arc ABC, find

.

In Fig. 10.38,

find

.

a triangle)

Therefore,

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

Solution

In

and

is the sum of the interior opposite angles.)

Therefore,

Question: 6

ABCD is a cyclic quadrilateral whose diagonals

intersect at a point E. If

=

Solution

In

,

(Angles in same segment of a circle

are equal.)

triangle)

below and its diagonals AC and BD be the diameters

of the circle through the vertices of the quadrilateral.

(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,

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

ABCD, such that,

(Right angles)

(Distance between the parallel lines)

by RHS congruence condition.

Now,

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

Solution

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

figure below.

ABD and PBQ are line segments intersecting at B.

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

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

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

prove that BDC is a straight line.

(Angle in the semi-circle)

Now,

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

Solution

Let

and