Circle with radii 3, 4, 5 touch each other externally. If Pis the point of intersection of tangents to these circles at their point of contact the distance of P from the point of contanct is

Circles with radii 3, 4, 5 touch each other externally and the point of intersection of tangents is P, then P is the incentre of $\mathrm{\Delta }{C}_1{C}_2{C}_3.$

So, distance of point P from the points of contact = lnradius of $\mathrm{\Delta }{C}_1{C}_2{C}_3$

and            $C_1{C}_2=7,C_2{C}_3=9,C_3{C}_1=8$

$\Rightarrow$                $a=7,b=9,c=8$

We know that   $r=\frac{\mathrm{\Delta }}{S}=\frac{\sqrt{S(S-a)(S-b)(S-c)}}{S}$

                   $\begin{array}{r}r=\sqrt{\frac{(s-a)(s-b)(s-c)}{s}}\\ =\sqrt{\frac{(12-7)(12-9)(12-8)}{12}}\end{array}$

                               $\left(\begin{array}{l}\because s=\frac{a+b+c}{2}=\frac{7+9+8}{2}\\ \therefore s=12\end{array}\right)$

$\therefore r=\sqrt{\frac{5\cdot 3\cdot 4}{12}}=\sqrt{5}$