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**Q.1Â Â Â Â DrawÂ differentÂ pairs of circles. HowÂ many pointsÂ does each pairÂ haveÂ in common? What is theÂ maximum number of common points? **

**Sol. **

On drawing a different pairs of circles, we find that they have two points in common. NowÂ we shall prove that a pair of circleÂ cannot intersect each other atÂ more than two points.

Let the twoÂ circles cut each other at three points. ButÂ from the threeÂ points only one circle passes. Hence, two circles if intersect each other then they will intersect at two points. These two points are the ends of their common chords. This chord can be a common chord of a number of circles which will pass through the end points of this common chord.

Â **Q.2Â Â Â Suppose you are given a circle. Give aÂ constructionÂ to findÂ its centre. **

**Sol.Â Â Â Â Â ****Steps of Construction : **

1.Â Â Take 3Â points A, B and C on theÂ circumference of the circle.

2.Â Â JoinÂ AB and BC.

3.Â Â Draw PQ and RS, the perpendicular bisectors of AB and BC, which intersect each other at O. Then,O is the centreÂ of the circle.

**Q.3Â Â Â Â If two circles intersec at twoÂ points, prove that theirÂ centres lie on the perpendicular bisector of the commonÂ chord. **

**Sol.Â **

**Given :** TwoÂ circles, withÂ centres O and O' intersect, at two points A and B so that AB is theÂ common chord of the two circles and OO' is the line segment joining the centres of the two circles. Let OO' intersect AB at M.

**To prove :** OO' is theÂ perpendicularÂ bisector of AB.

Construction : Draw line segment OA, OB, O'A and O'B.

**Proof :** OAO' and OBO' we have

OA = OB Â Â Â Â Â Â Â Â Â [Radii of the same circle]

O'A = O'B Â Â Â Â Â Â Â Â Â [Radii of the same circle]

and Â Â Â OO' = OO'Â Â Â Â Â Â

ByÂ SSS criterion ofÂ congruence, we have

Â Â

Â Â Â

Â Â Â Â Â Â Â Â Â ... (1)

[Since ]

In AOM and BOM,Â we have

OA = OB Â Â Â Â Â Â Â Â Â [Radii of the same circle]

Â Â Â Â Â Â [From (1)]

and Â Â Â OM = OM Â Â Â Â Â Â Â Â Â [Common]

ThereforeÂ by SAS criterion of congruence, we have

Â Â Â AM = BM and

ButÂ

Therefore

Thus , Â Â Â AM = BMÂ and

Hence , OO' is the perpendicularÂ bisector of AB.

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