## 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?**

Answer:

Answer:

Therefore, the maximum number of common points is 2.

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

Answer:

Answer:

**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.**

Answer:

Answer:

**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)

AD = BD (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.

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