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
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
14