Question

Two circles with centres O and O′ intersect at two points A and B. A line PQ is drawn parallel to OO′ through A(or B) intersecting the circles at P and Q. Prove that PQ = 2 OO′.

Answer 1

Firstly draw two circles with center O and O’ such that they intersect at A and B.

Draw a line PQ parallel to OO’.

In the circle with center O, we have:

OP and OB are the radii of the circle.

PB is the chord with OM as its perpendicular bisector.

i.e. BM = MP  ...(1)

In the circle with center O’, we have:

O’B and O’Q are the radii of the circle.

BQ is the chord with O’N as its perpendicular bisector.

i.e. BN = NQ  ...(2)

From (1) and (2), we have:

BM + BN = MP + NQ

⇒ (BM + BN) + (BM + BN) = (BM + BN) + (MP + NQ)

⇒ 2(BM + BN) = (BM + BN) + (MP + NQ)

⇒ 2(OO’) = (BM + MP) + (BN + NQ)

⇒ 2(OO’) = BP + BQ

⇒ 2(OO’) = PQ

Hence proved.