    Question

# Two circles whose centres are O and O′ intersect at P. Through P, a line parallel to OO′, intersecting the circle at C and D is drawn as shown. Then CD=2OO′. A

True

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B

False

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Solution

## The correct option is A True Construction: Draw perpendiculars OA and O′B to CD from O and O′, respectively. Since OA⊥CD, OA bisects the chord CP. (∵ Perpendicular from the centre of the chord bisects the chord) ∴ AP=12CP or CP=2 AP - - - - - (i) Similarly since O′B⊥PD, we must have BP=12PD or PD=2 BP - - - - - (ii) Now, CD=CP+DP ⟹CD=2AP+ABP=2(AP+BP) [from (i) and (ii)] ⟹CD=2(AB) - - - - - (iii) In quadrilateral ABO′O, OO′∥AB and ∠OAB=∠O′BA=90∘ So, OA∥O′B Thus ABO′O is a rectangle. ∴AB=OO′ Hence CD=2AB=2OO′ Therefore the given statement is true.  Suggest Corrections  28      Similar questions  Related Videos   Chord Theorem 1
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