Problem 4. (6 points) A circle with center $I$ is inside a circle with center $O$. In the larger circle, a chord $A B$ is constructed, which is tangent to the smaller circle. Determine the geometric locus of the center of the circumcircle of triangle $I A B$.

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Solution: We will prove that the sought locus is a circle with center $O$ and radius $\frac{R^{2}-d^{2}}{2 r}$, where $R$ and $r$ are the radii of the given circles, and $d=O I$.

Let $M$ be the midpoint of $A B$, $H$ be the foot of the perpendicular from $I$ to $O M$, and $X$ be the center of the circumcircle of $\triangle I A B$. Then

$$
\begin{aligned}
X B^{2} & =X M^{2}+M B^{2}=X M^{2}+R^{2}-O M^{2} \\
X I^{2} & =I H^{2}+X H^{2}=\left(d^{2}-(O M-r)^{2}\right)+(X M+r)^{2}
\end{aligned}
$$

But $X B=X I$ and from the above equalities we get $R^{2}=d^{2}+2 r(O M+X M)$, i.e., $O X=\frac{R^{2}-d^{2}}{2 r}$ does not depend on the choice of the chord $A B$. It is easily verified that all points on the circle with center $O$ and radius $O X$ belong to the sought locus.