On the plane, there are two concentric circles centered at point $A$. Let $B$ be an arbitrary point on one of these circles, and $C$ be a point on the other. For each triangle $A B C$, consider two circles of the same radius that touch each other at point $K$, where one circle touches the line $A B$ at point $B$, and the other touches the line $A C$ at point $C$. Find the locus of point $K$.
Let $M, N$ be the centers of touching circles. Then $K$ is the midpoint of segment $M N, \angle A B M=\angle A C N=90^{\circ}$, and $B M=M K=K N=N C$. Since $A K$ is the median of triangle $A M N, A K^{2}=\frac{2 A M^{2}+2 A N^{2}-M N^{2}}{4}=\left(A B^{2}+A C^{2}\right) / 2$ - does not depend on the choice of points $B, C$. Therefore, $K$ lies on a circle with center $A$. By rotating triangle $A B C$ around $A$, any point of this circle can be obtained.

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