NCERT Solutions For Class 9 Math Chapter -10 Exercise – 10.3

1. Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?

Solution:

Ncert solutions class 9 chapter 10-3

In these two circles, no point is common.

Ncert solutions class 9 chapter 10-4

Here, only one point “P” is common.

Ncert solutions class 9 chapter 10-5

Even here, P is the common point.

Ncert solutions class 9 chapter 10-6

Here,

two points are common which are P and Q.

Ncert solutions class 9 chapter 10-7

No point is common in the above circle.

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

Solution:

Ncert solutions class 9 chapter 10-8

The construction steps to find the center of the circle are:

Step I: Draw a circle first.

Step II: Draw 2 chords AB and CD in the circle.

Step III: Draw the perpendicular bisectors of AB and CD.

Step IV: Connect the two perpendicular bisectors at a point. This intersection point of the two perpendicular bisectors is the centre of the circle.

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

Solution:

Ncert solutions class 9 chapter 10-9

It is given that two circles intersect each other at P and Q.

To prove:

OO’ is perpendicular bisector of PQ.

Proof:

Triangle ΔPOO’ and ΔQOO’ are similar by SSS congruency since

OP = OQ and O’P = OQ (Since they are also the radii)

OO’ = OO’ (It is the common side)

So, It can be said that ΔPOO’ ΔQOO’

∴ POO’ = QOO’ — (i)

Even triangles ΔPOR and ΔQOR are similar by SAS congruency as

OP = OQ (Radii)

POR = QOR (As POO’ = QOO’)

OR = OR (Common arm)

So, ΔPOR ΔQOR

∴ PRO = QRO

Also, we know that

PRO+QRO = 180°

Hence, PRO = QRO = 180°/2 = 90°

So, OO’ is the perpendicular bisector of PQ.