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

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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

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

In triangle OQR,

∴ By Pythagoras Theorem,

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

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If d1,d2(d2>d1) be the diameters of two concentric circle and c be the length of a chord of a circle which is tangent to the other circle. prove that d22=c+d21.

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