## Chapter 11 Circles Ex 11.3

**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:**

Different pairs of circles are **(i) Two points common**

**(ii) One point is common**

**(iii) No point is common**

**(iv) No point is common**

**(v) One point is common**

From figures, it is obvious that these pairs many have 0 or 1 or 2 points in common.

Hence, a pair of circles cannot intersect each other at more than two points.

**Question 2.** **Suppose you are given a circle. Give a construction to find its centre.** **Solution:** **Steps of construction**

Taking three points P,Q and R on the circle.

Join PQ and QR,

Draw MQ and NS, respectively the perpendicular bisectors of PQ and RQ, which intersect each other at O.

Hence, O is the centre of the circle.

**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 with centres O and O’ intersect at two points M and N so that MN is the common chord of the two circles and OO’ is the line segment joining the centres of the two circles. Let OO’ intersect MN at P. **To prove:** OO’ is the perpendicular bisector of MN.

**Construction:** Draw line segments OM, ON, O’M and O’N.

Proof In ∆ OMO’ and ONO’, we get

OM = ON (Radii of the same circle)

O’M = O’N (Radii of the same circle)

OO’ = OO’ (Common)

∴ By SSS criterion, we get

∆ OMO’ ≅ ONO’

So, ∠ MOO’ = ∠ N00′ (By CPCT)

∴ ∠ MOP = ∠ NOP …(i)

(∵ ∠ MOO’ = ∠ MOP and ∠ NOO’ = ∠ NOP)

In ∆ MOP and ∆ NOP, we get

OM = ON (Radii of the same circle)

∠ MOP = ∠NOP [ From Eq(i)]

and OM = OM (Common)

∴ By SAS criterion, we get

∆ MOP ≅ ∆NOP

So, MP = NP (By CPCT)

and ∠ MPO = ∠ NPO

But ∠ MPO + ∠NPO = 180° ( ∵MPN is a straight line)

∴ 2 ∠ MPO = 180° ( ∵ ∠ MPO = ∠ NPO)

⇒ ∠ MPO = 90°

So, MP = PN

and ∠ MPO = ∠ NPO = 90°

Hence, OO’ is the perpendicular bisector of MN.