Question

Two circles with centers 0 and O' intersect at two points A and
1d 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.ro
or B) intersecting the circles at P and Q. Prove that PQ = 200

Answer 1

PQ  = 2OO' as OO'NM forms a rectangle

Centre of the circles O and O' (Given)

Intersection points of O' = A and B (Given)

Intersection points of O = P and Q (Given)

In △OPB,

BM = MP -- 1  ( As the perpendicular from center to circle bisects the chord)

Similarly in △O'BQ,  

BN = NQ  -- 2 ( As the perpendicular from center to circle bisects the chord)

Adding equation 1 and 2

= BM + BN = PM + NQ

Now, adding the equation BM + BN to both the sides

= BM + BN + BM + BN = BM + PM + NQ + BN

2BM + 2BN = PQ

2(BM + BN) = PQ

Now,    

OO = MN   [As OONM  forms a rectangle]

= PQ  = 2OO'

Hence proved.