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′.
Two circles with centres O and O′ intersect at two points A and B.
PQ || OO'
To Prove that: PQ = 2OO'
Proof:
Draw OM perpendicular to PB and O'N perpendicular to BQ
From the figure, we have:
OP = OB (radius of the same circle)
O'B = O'Q (radius of the same circle)
In OPB,
BM = MP … (i) (perpendicular from the centre of the circle bisects the chord)
Similarly in O'BQ,
BN = NQ … (ii) (perpendicular from the centre of the circle bisects the chord)
Adding (i) and (ii),
BM + BN = PM + NQ
Adding BM + BN to both sides.
BM + BN + BM + BN = BM + PM + NQ + BN.
2BM + 2BN = PQ
2(BM + BN) = PQ … (iii)
OO' = MN (As OO' NM is a rectangle)
OO' = BM + BN ... (iv)
Using equation (iv) in (iii)
2 OO'= PQ
Hence proved.