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\[{d_2}^2 = c^2 + {d_1}^2\].

#### 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,

\[\left( OQ \right)^2 = \left( OR \right)^2 + \left( RQ \right)^2 \]

\[ \Rightarrow \left( \frac{d_2}{2} \right)^2 = \left( \frac{d_1}{2} \right)^2 + \left( \frac{c}{2} \right)^2 \]

\[ \Rightarrow \left( d_2 \right)^2 = \left( d_1 \right)^2 + c^2\]