In the first part of our solution, we will assume that $X$ is any point with the required property. It is clear that $X$ lies in the exteriors of the circles $k_1$ and $k_2$ and that the points $S_1$, $S_2$ and $X$ are vertices of a triangle whose sides $S_1X$, $S_2X$ are intersected successively by the circles $k_1$ and $k_2$ in such points $Y_1$, $Y_2$ which lie on the same line parallel to $S_1S_2$ (Fig. 1). Since the triangles $XY_1Y_2$ and $XS_1S_2$ are similar (by theorem AA), it holds
$$\frac{|XY_1|}{|XS_1|} = \frac{|XY_2|}{|XS_2|}, \qquad (1)$$
which can be rewritten, because of the equalities
$$|XY_1| = |XS_1| - r_1 \quad \text{a} \quad |XY_2| = |XS_2| - r_2, \qquad (2)$$
as an equation for lengths of the segments $XS_1$ and $XS_2$:
$$\frac{|XS_1| - r_1}{|XS_1|} = \frac{|XS_2| - r_2}{|XS_2|}, \\ \frac{|XS_1|}{|XS_2|} = \frac{r_1}{r_2}. \qquad (3)$$
Since the points $S_1$ and $S_2$ are fixed as well as the ratio $r_1/r_2$, the locus of points $X$

Fig. 1

satisfying equation (3) is a circle of Apollonius (which evidently becomes a straight line if the ratio $r_1/r_2$ equals 1). As known for the case when $r_1/r_2 \neq 1$, there are exactly two solutions $X = H_1$ and $X = H_2$ of the equation (3) that lie on the line $S_1S_2$ and form a diameter $H_1H_2$ of the resulting circle of Apollonius. For our situation, let us add the fact that the points $H_1$ and $H_2$ are centres of both homotheties of the initially given circles $k_1$ and $k_2$.

In the second part of our solution, we will conversely assume that $X$ is a point of the circle of Apollonius given by the equation (3) and that $X$ lies outside of the line $S_1S_2$, i.e. $X \neq H_1$ and $X \neq H_2$. In view of the condition that $|S_1S_2| > r_1 + r_2$, the whole circle of Apollonius (with diameter $H_1H_2$) lies in the exteriors of the circles $k_1$ and $k_2$. Indeed, the last fact follows from a known position of the homothety centres $H_1$ and $H_2$ on the line $S_1S_2$: If for example $r_1 < r_2$, then the diameter $H_1H_2$ contains the diameter of $k_1$, while the diameter of $k_2$ and the diameter $H_1H_2$ are disjoint. (See also *Remark* below.)

The proven property implies that $XS_1S_2$ is a triangle with $|XS_1| > r_1$ and $|XS_2| > r_2$. Thus there are points $Y_1 \in k_1$ and $Y_2 \in k_2$ lying on the segments $S_1X$ and $S_2X$, respectively. Since the equalities (2) are valid again, it is possible to transform the equation (3) to equation (1). Consequently, the triangles $XS_1S_2$ and $XY_1Y_2$ are similar (now by SAS theorem) and hence $S_1S_2 \parallel Y_1Y_2$. Therefore the distances of $Y_1$ and $Y_2$ to the line $S_1S_2$ are equal which proves the required property of the point $X$.

*Answer.* If $r_1 \neq r_2$, the locus of points $X$ is the circle of Apollonius which is given by the above equation (3), excepting the two points on the line $S_1S_2$. If $r_1 = r_2$, the locus is the perpendicular bisector of the segment $S_1S_2$, with exception of the midpoint of $S_1S_2$.

Fig. 1