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## 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? Therefore, the maximum number of common points is 2.

Question 2: Suppose you are given a circle. Give a construction to find its centre. Construction steps:
1. Trace a circle through any circular object.
2. Mark three 3 non-collinear points on the circle - A, B and C and join them.
3. Draw the perpendicular bisectors of AB and AC.
4. We observe that the perpendicular bisectors intersect at a point (O).
5. We know that every point on the perpendicular bisector of a line segment is equidistant from the endpoints.
6. Therefore the point of intersection (O) of the perpendicular bisectors indicates 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. Given:
Circles with centres C and D (intersecting circles). AB is the common chord.

Required to prove:
CD is the perpendicular bisector of AB.

Proof:
Join AC, BC, BD, AD. Let AB and CD intersect at O.

Considering ∆ACD and ∆BCD
CD = CD         (common)
AC = BC         (radius of the same circle)

Therefore by SSS congruency, ∆ACD ≅ ∆BCD.
By CPCT ∠ACD = ∠BCD.

Again considering ∆ACO and ∆BCO
CO = CO                     (common)
AC = BC                     (radius of the same circle)
∠ACD = ∠BCD          (radius of the same circle)

Therefore by SAS congruency, ∆ACO ≅ ∆BCO.
By CPCT, AO = BO and ∠AOC = ∠BOC.

But ∠AOC = ∠BOC form a linear pair.
∠AOC + ∠BOC = 180°
2∠AOC = 180°
∠AOC = 90°

Hence CD bisects AB perpendicularly.