Question

23. Two circles with centres O and O' intersect at two points A and B. A line
PQ is drawn parallel to 00' through A or B, intersecting the circles at P
and Q. Prove that PQ = 200'.​

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....(1)BM=MP....(1)

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⇒2OO’=PQ

Hence, proved.