10.255. Through point $A$ of a circle with a radius of 10 cm, two mutually perpendicular chords $A B$ and $A C$ are drawn. Calculate the radius of the circle that is tangent to the given circle and the constructed chords, if $A B=16$ cm.
Solution.

$O$ is the center of a circle with a radius of 10 cm (Fig. 10.60), $O_{1}$ is the center of a circle that touches the chords $A B$ and $A C$ and the given circle at points $M, D$, and $K$ respectively. Then $O_{1} D \perp A C, O_{1} M \perp A B$. $\angle B A C$ is an inscribed and right angle. Therefore, $B C$ is the diameter of the given circle, $B C=20 \text{ cm}$, $A C=\sqrt{B C^{2}-A B^{2}}=\sqrt{20^{2}-16^{2}}=12$ (cm). Drop perpendiculars $O L$ and $O N$ to $A C$ and $A B$ respectively. Then $A L=L C=\frac{1}{2} A C=6$ cm, $A N=N B=\frac{1}{2} A B=8$ cm. Let the desired radius $O_{1} K=O_{1} D=O_{1} M=$ $=x$ cm. $E$ is the point of intersection of $O L$ and $O_{1} M$. Then $\angle O E O_{1}=90^{\circ}$, $O E=O L-E L=A N-O_{1} D=(8-x) \text{ cm}, Q_{1} E=O_{1} M-M E=O_{1} M-A L=(x-6) \text{ cm}$.
$O Q_{1}=O K-Q_{1} K=(10-x)$ cm. From $\triangle O E Q\left(\angle O E Q=90^{\circ}\right): O E^{2}+E O_{1}^{2}=O O_{1}^{2} ;$ $(8-x)^{2}+(x-6)^{2}=(10-x)^{2} ; x=8$.

Answer: 8 cm.