These NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.3 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 9 Maths Chapter 10 Circles 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:

There are three different pairs of circle.

It is clear that in Fig. I, there is no any common point in pair of circle O and O’.

In Fig. II, there is only one point A is common point in pair of circle P and P’.

In Fig. III, there are two points X and Y are common points in pair of circle Q and Q’.

The maximum number of common points are two.

Question 2.

Suppose you are give a circle. Give a construction to find its centre.

Solution:

To find the centre of the circle we take following steps:

Step (i) – Take three points A, B and Con die circle.

Step (ii) – Join AB and AC.

Step (iii) – Draw perpendicular bisector of AB and BC.

Step (iv) – Both perpendicular bisector intersect each other at point O.

Step (v) – Point O is centre of the required circle.

Remark: We know that centre lie on perpendicular bisector of chord. Therefore centre must be lie on perpendicular bisector of AB. Again centre also must be lie on perpendicular bisector of BC. Both cases are possible only when centc lie on their point of intersection.

Question 3.

If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.

Solution:

Given: Two circles O and O’ intersect each other at A and B. AB is a common chord for both circles.

To prove that: OO’ is perpendicular bisector of AB.

Construction: Draw line segments OA, OB, O’A and O’B.

Proof: In ∆OAO’ and ∆OBO’ we have

OA = OB (Radii of same circle)

O’A = O’B (Radii of same circle)

and OO’ = OO’ (Common)

So, by S-S-S congruency condition

∆OAO’ ≅ ∆OBO’

∴ ∠AOO’ = ∠BOO’ (By C.P.C.T)

or ∠AOM = ∠BOM …..(i)

Now, In ∆AOM and ∆BOM

OA = OB (Radii of samecircle)

∴ ∠AOM = ∠BOM (From (i))

OM = OM (Common)

So, by S-A-S congruency condition

∆AOM ≅ ∆BOM

∴ AM = BM (By CPCT)

and ∠AMO = ∠BMO (By CPCT)

But, ∠AMO + ∠BMO = 180° (Liner pair)

2∠AMO = 180° (∵ ∠AMO = ∠BMO prove above)

or, ∠AMO = 90°

Hence, OO’ lie on the perpendicular bisector of AB.