If three circles with radii ${3}$,${4}$,${5}$ touch each other externally at points P,Q and R,then the CIRCUMRADIUS of ∆PQR is...??

My attempt i think that the let the point of the common intersection of the tangents to these 3 circles be A then A will be the $(circumcentre)$ of the ∆PQR??


Let $A,B,C$ be the centers of the circles with radii $u=3,v=4,w=5$ resp.

The sides of this triangle are $a=v+w=9, b=u+w=8, c=u+v=7$.

It is known that these equations are only fulfilled iff points PQR are the tangency points of the incircle of triangle ABC on its sides.

The radius $r$ of the incircle is known to be given by the following formula:

$$r^2=\dfrac{1}{s}(s-a)(s-b)(s-c) \ \ \ \text{where} \ \ \ s=\dfrac{a+b+c}{2}$$


thus, with $s=12$, $r^2=\frac{60}{12}=5$, therefore $r=\sqrt{5}$.


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