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Exercise 10.3

1. Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?

Sol.

(i)
NO point is common

(ii)
One point P is Common.

(iii)
One point P is common.

(iv)
Two points P and Q are common.
Hence, the maximum number of common points is two, which is incase (iv).

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

Sol. Steps to find centre of the circle:

Two non-parallel chords AB and CD of a circle are drawn.
(ii) Perpendicular bisectors of AB and CD are drawn.
(iii) Let these bisectors meet at O.
Then 0 is the required centre of the circle.

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

Sol. Let two circles with centres 0 and 0′ intersect each other at P and Q. Thus PQ is the common chord as shown in the adjoining figure.

Let us draw perpendicular OL on PQ,
then OL bisects PQ at L,

[Perpendicular from centre of a circle to the chord bisects the chord]

i.e., ∠OLP = 90° and PL = QL                                    ……(i)

L and 0′ are joined

Then O’L is perpendicular to PQ                                 …(ii)

[Line segment joining the centre of the circle to midpoint of the chord is perpendicular to the chord]

From (i) and (ii), we have

∠OLP + ∠O’LP = 90° + 900 = 180′

⇒  ∠OLP and ∠O’LP from a linear pair.

Hence, OLO’ is a straight line

with PL = QL.

Hence centres of the two circles lie on the perpendicular bisector of the common chord.