NCERT Class 9 Maths Chapter 10 Circles Exercise 10.3

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:


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:


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:

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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