Let's modify the following curves.

$$
x^{2}+y^{2}=225, \quad x^{2}-30 x+y^{2}+189=0
$$

Construct the common tangents of these curves, determine the equations of these tangents and the coordinates of the points of tangency.
The first equation is the equation of a circle whose center $O$ is the origin of the coordinate system, and whose radius is 15. The second equation can also be written as:

$$
(x-15)^{2}+y^{2}=6^{2}
$$

We see that this equation is for a circle whose radius is $6$, and the distance of its center $O_{1}$ from $O$ is 15. Since the circles intersect each other, there are only two common tangents. Let the points of tangency be $M$ and $M_{1}$, and the tangents be $m$ and $m_{1}$, respectively, and the intersection point of the tangents be $A$; the angles $\angle AOM = \angle AO_{1}m = \alpha$, then

$$
\left(O O_{1}+O_{1} A\right): O_{1} A=O M: O_{1} m
$$

from which, after substituting the values,

$$
O_{1} A=10
$$

The equation of the tangent is of the form

$$
x \cos \alpha+y \sin \alpha=O M
$$

but

$$
\cos \alpha=\frac{O M}{O A}=\frac{3}{5} \text { and } \sin \alpha=\frac{M A}{O A}=\frac{4}{5}
$$

so the equation of the tangent is

$$
3 x+4 y=75
$$

The equation of the other tangent is

$$
3 x-4 y=75
$$

Since the points of tangency satisfy both the circle and the tangent equations, the coordinates of point $M$ are the solutions of $x^{2}+y^{2}=225$ and equation (2), which are $(9,12)$ and thus the coordinates of point $M_{1}$ are $(9,-12)$; the coordinates of point $m$ are the solutions of equations (1) and (2), which are $\left(18 \frac{3}{5}, 4 \frac{4}{5}\right)$ and thus the coordinates of point $m_{1}$ are $\left(18 \frac{3}{5},-4 \frac{4}{5}\right)$.

(Kázmér Smodics, Veszprém.)

The problem was also solved by: Bayer B., Czank K., Filkorn J., Kerekes T., Krausz B., Krisztián Gy., Lukhaub Gy., Lupsa Gy., Perl Gy., Póka Gy., Sasvári G., Selényi M.