Problem 728

If the given circles have a common tangent $s$, intersecting the line $m$ at point $S$, then we can consider that point $S$ satisfies the conditions of the problem, as through this point pass two coinciding tangents to the given circles equally inclined to the line $m$.
Ignoring this trivial solution, we will look for other solutions. Suppose that through point $T$ of the line $m$ pass two different lines $t_1$ and $t_2$, tangent respectively to circles $C_1$ and $C_2$ and equally inclined to the line $m$, i.e., symmetric with respect to $m$ (Fig. 26). The circle $C$, symmetric to the circle $C_1$ with respect to $m$, is tangent to the line symmetric to the tangent $t_1$ of the circle $C_1$, i.e., to the line $t_2$. Therefore, the line $t_2$ is a common tangent of the circles $C$ and $C_2$. The solution to the problem is the point of intersection $T$ of each common tangent of the circles $C$ and $C_2$ with the line $m$. Note that point $T$ is also the point of intersection with the line $m$ of the common tangent $t_1$ of the circle $C_1$ and the circle $C$ symmetric to the circle $C_2$ with respect to the line $m$. The problem can have $4$, $3$, $2$, $1$, or $0$ solutions.