**Exercise: 10.3 (Page No: 176)**

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

**Solution:**

In these two circles, no point is common.

Here, only one point āPā is common.

Even here, P is the common point.

Here, two points are common which are P and Q.

No point is common in the above circle.

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

**Solution:**

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

**Step I:**Ā Draw a circle first.

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

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

**Step IV:**Ā 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:**

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

**To prove:**

OOā is perpendicular bisector of PQ.

**Proof:**

Triangle ĪPOOā and ĪQOOā are similar by SSS congruency since

OP = OQ and OāP = OQ (Since they are also the radii)

OOā = OOā (It is the common side)

So, It can be said that ĪPOOā ĪQOOā

ā“ POOā = QOOā ā (i)

Even triangles ĪPOR and ĪQOR are similar by SAS congruency as

OP = OQ (Radii)

POR = QOR (As POOā = QOOā)

OR = OR (Common arm)

So, ĪPOR ĪQOR

ā“ PRO = QRO

Also, we know that

PRO+QRO = 180Ā°

Hence, PRO = QRO = 180Ā°/2 = 90Ā°

So, OOā is the perpendicular bisector of PQ.