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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 July 2017, 1:27 PM
Answered by 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 Expert 27th July 2017, 3:27 PM
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