If $d_{1}$ and $d_{2}$ ( $d_{2}$ > $d_{1}$ ) are the diameters of two concentric circles and C is the length of a chord of a circle which is tangent to the order circle, prove that $d_{2}^{2} = \frac{c^{2}}{4} + d_{1}^{2}$ .

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Solution

The question seems to be wrong. The question should have been "Prove that $d_{2}^{2} = d_{2}^{1} + C^{2}$
$A B = \frac{c}{2}$ [As c $\to$ tangent to the inner circle
$O A ⊥ A B$ at A and it divides the chord of length C into 2 equal half]
In $Δ O A B$ ,
By Pythagoras theorem,
$O B^{2} = O A^{2} + A B^{2}$
${(\frac{d_{2}}{2})}^{2} = {(\frac{d_{1}}{2})}^{2} + {(\frac{c}{2})}^{2}$
$∴ d_{2}^{2} = d_{1}^{2} + c^{2}$ .

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