10,11

Construct a circle equidistant from four points on a plane. How many solutions does the problem have?
Answer: 7 solutions (in the non-degenerate case). Let $A, B, C, D$ be the given points, and $S$ be the sought circle. On one side of $S$, there are $k$ given points, and on the other side, there are 4 - $k$ given points. We will assume that the given points do not lie on the same circle (otherwise, $S$ can be any circle with the same center, resulting in infinitely many solutions). Thus, $1 \leq k \leq 3$. We obtain two significant
perpendicular bisectors of segments $A B$ and $C D$. The radius of circle $S$ is the arithmetic mean of the lengths of segments $O A$ and $O C$. The four points can be paired in three ways, so we get 3 solutions. Now let points $A, B$, and $C$ lie on one side of circle $S$, and point $D$ lie on the other side. Draw a circle through points $A, B$, and $C$. Let $O$ and $R$ be its center and radius. Point $O$ is the center of the sought circle, and the radius of the sought circle is the arithmetic mean of $R$ and $O D$. One point out of four can be chosen in four ways, so we get 4 solutions.