Q. 14

Two equal circles of radius r intersect such that each passes through the centre of the other. The length of the common chord of the circles is
A.

B. r AB

C. r

D. r

Answer :

Let O and O' be the centre of two circles

OA and O'A = Radius of the circles


AB be the common chord of both the circles


OM perpendicular to AB


And,


O'M perpendicular to AB


AOO' is an equilateral triangle.


AM = Altitude of AOO'


Height of AOO' = r


AB = 2 AM


= 2 r


= r

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