Circle: Common tangents of circles

Given : AB and MN are common tangents of circles O and P . AB = 15 and MN = 5 .
Find : the product of $\displaystyle {{r}_{1}}\ and\ {{r}_{2}}$ .

Solution

NM produced meet AB at C .

Join OA , OC , OP , PC , PB and draw $\displaystyle OD\bot PB$ .

$\displaystyle \text{Let}\ AC=CM=x$

$\displaystyle \ \ \ \ \ \ CN=CB\ \ (\ \because \ \text{tangents}\ )$

$\displaystyle \ \ \ \ \ \ x+5=15-x$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ x=5$

$\displaystyle \ \ \ 2\theta +2\beta ={{180}^{\circ }}$

$\displaystyle \ \ \ \ \ \ \theta +\beta ={{90}^{\circ }}$

$\displaystyle \text{In}\ \text{rt}\ \Delta OCP\ ,\ O{{P}^{2}}=O{{C}^{2}}+C{{P}^{2}}$

$\displaystyle \text{In}\ \text{rt}\ \Delta ODP\ ,\ O{{P}^{2}}=O{{D}^{2}}+D{{P}^{2}}$

$\displaystyle \ \ \ \ \ O{{C}^{2}}+C{{P}^{2}}=O{{D}^{2}}+D{{P}^{2}}$

$\displaystyle \ ({{r}_{1}}^{2}+{{5}^{2}})+({{r}_{2}}^{2}+{{10}^{2}})={{15}^{2}}+{{({{r}_{2}}-{{r}_{1}})}^{2}}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 125=225-2{{r}_{1}}{{r}_{2}}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2{{r}_{1}}{{r}_{2}}=100$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \therefore \ {{r}_{1}}{{r}_{2}}=50$