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 = 2OO'.

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Solution

Given: Two circles with centres O and O' intersect at two points A and B.

Draw a line PQ parallel to OO' through B, OX perpendicular to PQ, O'Y perpendicular to PQ, join all.

We know that perpendicular drawn from the centre to the chord, bisects the chord.

∴ PX = XB and YQ = BY

∴ PX + YQ = XB + BY

On adding XB + BY on both sides, we get

PX + YQ + XB + BY = 2(XB + BY)
⇒ PQ = 2(XY)
⇒ PQ = 2(OO')

Hence, PQ = 2OO'.

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A

and

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Question 18
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 whose centres are $O$ and $O^{'}$ intersect at $P$ . Through $P$ , a line parallel to $O O^{'}$ , intersecting the circle at C and D is drawn as shown. Then $C D = 2 O O^{'}$ .

Q. In the following figure: P and Q are the points of intersection of two circles with centres O and O’. If straight lines APB and CQD are parallel to OO’. Prove that
(i) OO’ = ½ AB
(ii) AB = CD

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