If $d_{1}, d_{2} (d_{2} > d_{1})$ be the diameters of two concentric circle s and c be the length of a chord of a circle which is tangent to the other circle , prove that ${d_{2}}^{2} = c^{2} + {d_{1}}^{2}$ .

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Solution

Let O be the centre of two concentric circles and PQ be the tangent to the inner circle that touches the circle at R.

Now, OQ= $\frac{1}{2} d_{2}$ and OR= $\frac{1}{2} d_{1}$

Also, PQ = c

As, PQ is the tangent to the circle.

⇒ OR ⊥ PQ

⇒ QR = $\frac{1}{2} PQ = \frac{1}{2} c$

In Triangle OQR,

∴ By Pythagoras Theorem,

${(O Q)}^{2} = {(O R)}^{2} + {(R Q)}^{2} \Rightarrow {(\frac{d_{2}}{2})}^{2} = {(\frac{d_{1}}{2})}^{2} + {(\frac{c}{2})}^{2} \Rightarrow {(d_{2})}^{2} = {(d_{1})}^{2} + c^{2}$

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