Question

Prove that the line joining the centres
of two intersecting circles is the
perpendicular bisector of the line
joining the points of intersection.​

Answer 1

Let A and B be centres of two circles intersecting at points P and Q.

In △APB and △AQB,

AP=AQ∵They are radii of a circle.

BP=BQ∵They are radii of a circle.

AB=AB∵Common side

△APB≅△AQB By S.S.S. Criterion

⇒∠PAB=∠QAB By C.P.C.T.C.

Now, In △APR and △AQR

AP=AQ∵ They are radii of a circle.

∠PAR=∠QAR∵∠PAB=∠QAB

AR=AR∵ Common side

△APR≅△AQR ....... By S.A.S. Criterion

⇒PR=RQ ..... By CPCT

And ∠ARP=∠ARQ ....... By C.P.C.T.C.

Also, ∠ARP+∠ARQ=180°

∵ PQ is a straight line.

⇒∠ARP+∠ARP=180°

⇒2×∠ARP=180°

⇒∠ARP= 180°/2 =90°

Thus, AB⊥PQ

Answer 2

Answer:

A AB is the common chord of the two circles and OO' is the line segment joining the centres of the two circles. Let OO' intersect AB at P. To prove: OO' is the perpendicular bisector of AB. Hence, OO' is the perpendicular bisector of AB.