[ Tangent Circles ]

Circle $S$ with its center at the vertex of the right angle of a right triangle is tangent to the circle inscribed in this triangle. Find the radius of circle $S$, given that the legs of the triangle are 5 and 12.

#
Let the circle $S$ with center at the vertex $C$ of the right angle of the right triangle $ABC$ with legs $AC=12$, $BC=5$, and hypotenuse $AB=13$ touch the inscribed circle of this triangle externally at point $K$ (Fig.1).

Let $O$ be the center of the inscribed circle of the given triangle, and $M$ be the point of tangency of the inscribed circle with the leg $BC$. Then

$$
OM=\frac{1}{2}(AC+BC-AB)=\frac{1}{2}(12+5-13)=2
$$

The line of centers of the touching circles passes through the point of tangency, so $OC=CK+OK$, $\angle OCM=45^{\circ}$, since $CO$ is the bisector of angle $ACB$. From the right triangle $OCM$, we find that $OC=\sqrt{2} OM=2\sqrt{2}$, and since $OK=OM$, we have $2\sqrt{2}=2+CK$. Therefore, $CK=2\sqrt{2}-2$, i.e., the radius of circle $S$ is $2\sqrt{2}-2$.

If the circle $S$ touches the inscribed circle of the given triangle internally (Fig.2), then similarly we find that the radius of circle $S$ is $2\sqrt{2}+2$.

## Answer

$2(\sqrt{2} \pm 1)$