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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′.

Answers (1)

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 \triangleOPB,

BM = MP        … (i)   (perpendicular from the centre of the circle bisects the chord)

Similarly in \triangleO'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.

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