Question

The two circles in the picture cross each other at A and B. The points P and Q are the

i. Prove that P, B, Q lie on a line. 

ii. Prove that PQ is parallel to the line joining the centres of the circles and is twice as long as this line.

Answer 1

i. ∆ PAB (angle subtended by semicircle) ∠PBA = 90°

∆ ABQ be a triangle on the semi¬circle of centre D.

∴ ∠ABQ = 90°

∠PBA + ∠ABQ = 180 (Linear pair)

As AP, AQ are diameters of the circle. PQ be the line drawn through B per¬pendicularly to AB. 

Therefore P, B, Q lies on the same line.

ii.

Let M and N be the centre of the circles.

AP = 2 AM; \(\frac{AP}{AM}=2\) Similarly,

\(\frac{AQ}{AN}=2;\) ∠MAN = ∠PAQ (common angle)

△AMN \(\sim\) △PAQ As the triangles are similar, ∠AMN = ∠APQ.

∴ MN || PQ, PQ = 2 MN.