Given point $M$.

Given are two intersecting circles with centers $O_{1}, O_{2}$. Construct a circle that touches one of them externally and the other internally, the center of which is at the greatest distance from the line $O_{1} O_{2}$.
Let $O, r$ be the center and radius of a certain circle tangent to the given ones; $r_{1}, r_{2}$ be the radii of the given circles. Then either $O O_{1}=r_{1}-r, O O_{2}=r_{2}+r$, or $O O_{1}=r_{1}+r, O O_{2}=r_{2}-r$, and in both cases $O O_{1}+O O_{2}=r_{1}+r_{2}$. Therefore, among all points satisfying this condition, we need to find the one farthest from the line $\mathrm{O}_{1} \mathrm{O}_{2}$. The greatest height among all triangles with one given side and the sum of the other two sides is that of an isosceles triangle (see the solution to problem $\underline{5613}$). From this, we obtain that the center of the desired circle lies at equal distances $1 / 2\left(r_{1}+r_{2}\right)$ from points $O_{1}$ and $O_{2}$, and its radius is $1 / 2\left|r_{1}-r_{2}\right|$.