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.

Asked by Topperlearning User | 27th Jul, 2017, 01:27: PM

Expert Answer:

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,

OC2+BC2=OB2

i.e. , (as C bisects AB)

Therefore, d22= c2+d12

Answered by  | 27th Jul, 2017, 03:27: PM