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

 Solution:

It is given that two circles intersect each other at P and Q.

To prove:

OO’ is perpendicular bisector of PQ.

Proof:

Triangle ΔPOO’ and ΔQOO’ are similar by SSS congruency since

OP = OQ and O’P = OQ (Since they are also the radii)

OO’ = OO’ (It is the common side)

So, It can be said that ΔPOO’ ≅ ΔQOO’

∴ ∠POO’ = ∠QOO’ — (i)

Even triangles ΔPOR and ΔQOR are similar by SAS congruency as

OP = OQ (Radii)

∠POR = ∠QOR (As ∠POO’ = ∠QOO’)

OR = OR (Common arm)

So, ΔPOR ≅ ΔQOR

∴ ∠PRO = ∠QRO

Also, we know that

∠PRO+∠QRO = 180°

Hence, ∠PRO = ∠QRO = 180°/2 = 90°

So, OO’ is the perpendicular bisector of PQ.