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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