Two circles with centers 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. Then PQ = 3 OO′.

Creating figure using given data:

Considering triangle OPB:
From figure OM is the perpendicular bisector of

P B

from the center O, that bisects the chord.
So,

In triangle OBQ,

O ′ N

is the perpendicular bisector of BQ drawn from the center

O ′,

that bisects that chord

B Q

.
So,

B N = N Q ...... 2

Adding equation 1 and equation 2.

B M + B N = M P + N Q

Adding

B M + B N

on both sides,

B M + B N + B M + B N = B M + B N + M P + N Q

From the figure we have,

O O ′ = M N = M B + B N

B M + M P = B P

B N + N Q = B Q

By substituting the result obtained above in equation 3 we get,

2 O O ′ = B P + B Q

So, the statement is false,

2 O O ′ = B P + B Q

.
Hence option 2 is correct.

JEE