Circles with radii 3,4,5 are touching each other externally. If P is the point of intersection of tangents to these circles at their points of contact, the distance of P from the point of contact is

$$\begin{gathered} {\mathrm{r}}_1=3, {\mathrm{r}}_2=4, {\mathrm{r}}_3=5 \\ \mathrm{Let} \mathrm{A},\mathrm{B},\mathrm{C} \mathrm{are} \mathrm{centres} \mathrm{of} \mathrm{circles} \mathrm{respectively} \mathrm{we} \mathrm{know} \\ \mathrm{that} \mathrm{AP} \mathrm{bi}-\mathrm{sects} \mathrm{angle} \mathrm{formed} \mathrm{by} \mathrm{the} \mathrm{sector} \mathrm{at} \mathrm{centre} \mathrm{A}. \\ \mathrm{Similarly} \mathrm{CP} \mathrm{and} \mathrm{BP}. \mathrm{the} \mathrm{point} \mathrm{p} \mathrm{is} \mathrm{the} \mathrm{Incentre} \mathrm{of} ∆\mathrm{ABC}. \\ \mathrm{r}=\frac{∆}{\mathrm{s}}=\frac{\sqrt{\mathrm{s}(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})}}{\mathrm{s}} \\ 2\mathrm{s}=9+8+7=24 \therefore \mathrm{s}=12 \\ \mathrm{r}=\sqrt{\frac{5\times 4\times 3}{12}}=\sqrt{\frac{60}{12}}=\sqrt{5} \end{gathered}$$