# Circle

## Exercise 10.2

Question 1: Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres.

Answer: A per the theorem; equal chords (of a circle) subtend equal angles at the centre. Hence, it is clear that equal chords of congruent circles would subtend equal angles at their centres.

Question 2: Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal.

Answer: This is same as the previous question. The theorem says that if two chords subtend equal angles at centre then the chords are equal. Hence, if two chords in two congruent circles subtend equal angles at their centres then the chords are equal.

### 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: The maximum number of common points between a pair of circles can be two. Thus, it can also be zero or 1.

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

Answer: Let us assume that an arc PQ is given; as shown in this figure.

Let us take a point R on the arc.
Join the point PR and PQ.
Draw perpendicular bisectors of PR and PQ.
It is observed that perpendicular bisectors of chords PR and PQ intersect at O.
Taking O as centre and OR as radius, draw the circle.
It is observed that the circle passes through P, R and Q.
This proves that through any three non-collinear points; one and only one circle can pass.

Question 3: If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.

Answer: Given two circles intersect at points A and B.
To Prove: AM = MB
∠AMO = ∠BMO = 90°

In triangles AMO and BMO;
AO = BO (radii of circle)
OM = OM (common side)
Hence, from SSS theorem; ΔAMO ≈ ΔBMO
So, AM = MB proved
Moreover, ∠AMO = ∠BMO = 90° proved.

Hence, it is proved that when two circles intersect at two points then their centres lie on the perpendicular bisector of the common chord.

## Exercise 10.4

Question 1: Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.

Answer: Let us assume that OA = 5 cm, O’A = 3 cm and OO’ = 4 cm then AB = ?

In ΔAMO; AM2 = OA^2 - OM^2
Or, AM^2 = 5^2 - OM^2
Or, AM^2 = 25 – OM^2 ……………(1)

In ΔAMO’; AM^2 = OA’^2 - (4 – OM)^2
AM^2 = 3^2 - (4^2 + OM^2 - 8OM)
= 9 – 16 – OM^2 + 8OM ………….(2)

From equations (1) and (2);
25 – OM^2 = 8OM – OM^2 - 7
Or, 8OM = 25 + 7 – OM^2 + OM^2
Or, 8OM = 32
Or, OM = 32/8 = 4 cm
This means that O’M = 0

Using the value of OM, value of AM can be calculated as follows:
AM^2 = 25 – 4^2 = 25 – 16 = 9
Or, AM = 3 cm
Hence, chord AB = 2 xx 3 = 6  cm