If d1,d2 (d2>d1) are the diameters of two concentric circles and c is
the length of a chord of a circle which is tangent to the other circle,
then prove that d22= c2+d12.
Let AB = c be a chord of the larger circle, of diameter d2, which
touches the other circle at C. Then ∆OCB is a right triangle.
By Pythagoras theorem,
i.e. , (as C bisects AB)
Therefore, d22= c2+d12
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