What is the geometric place of the intersection points of the perpendicular tangents drawn to the circle $x^{2}+y^{2}=32$?
Let the equation of a circle be:

$$
x^{2}+y^{2}=r^{2}
$$

then the equation of the tangent at the point $(\xi, \eta)$ of the circle is:

$$
x \xi+y \eta=r^{2}
$$

If this tangent passes through the point $P\left(x_{0}, y_{0}\right)$, then

$$
x_{0} \xi+y_{0} \eta=r^{2}
$$

and since by assumption

$$
\xi^{2}+\eta^{2}=r^{2}
$$

therefore, from (3) and (4), $\xi, \eta$ can be determined, and we find that generally two values are obtained for both $\xi$ and $\eta$, which means that from any point $P$ generally two tangents can be drawn to the circle.

The result of the calculation is that

$$
\xi=\frac{r}{x_{0}^{2}+y_{0}^{2}}\left(r x_{0} \pm y_{0} \sqrt{x_{0}^{2}+y_{0}^{2}-r^{2}}\right)
$$

and

$$
\eta=\frac{r}{x_{0}^{2}+y_{0}^{2}}\left(r y_{0} \pm x_{0} \sqrt{x_{0}^{2}+y_{0}^{2}-r^{2}}\right)
$$

The equations of the two tangents passing through the point $P$ can now be obtained by substituting the values of $\xi_{1}, \eta_{1}$ and $\xi_{2}, \eta_{2}$ into equation (2), where we find that

$$
t_{1} \equiv y=-\frac{r x_{0}+y_{0} \sqrt{x_{0}^{2}+y_{0}^{2}-r^{2}}}{r y_{0}+x_{0} \sqrt{x_{0}^{2}+y_{0}^{2}-r^{2}}} \cdot x+\frac{r\left(x_{0}+y_{0}^{2}\right)}{r y_{0}+x_{0} \sqrt{x_{0}^{2}+y_{0}^{2}-r^{2}}}
$$

and

$$
t_{2} \equiv y=-\frac{r x_{0}-y_{0} \sqrt{x_{0}^{2}+y_{0}^{2}-r^{2}}}{r y_{0}-x_{0} \sqrt{x_{0}^{2}+y_{0}^{2}-r^{2}}} \cdot x+\frac{r\left(x_{0}^{2}+y_{0}\right)}{r y_{0}-x_{0} \sqrt{x_{0}^{2}+y_{0}-r^{2}}}
$$

If $t_{1}$ and $t_{2}$ are perpendicular to each other, then the product of their slopes is -1, so

$$
\frac{r x_{0}+y_{0} \sqrt{x_{0}^{2}+y_{0}^{2}}-r^{2}}{r y_{0}+x_{0} \sqrt{x_{0}+y_{0}^{2}-r^{2}}} \cdot \frac{r x_{0}-y_{0} \sqrt{x_{0}^{2}+y_{0}^{2}}-r^{2}}{r y_{0}-x_{0} \sqrt{x_{0}^{2}+y_{0}^{2}-r^{2}}}=-1
$$

from which

$$
\left(x_{0}+y_{0}^{2}\right)\left(x_{0}^{2}+y_{0}^{2}-2 r^{2}\right)=0
$$

If we assume that $P$ is not at the origin, then

$$
x_{0}+y_{0}^{2}=\overline{O P}^{2} \neq 0
$$

so it must be that

$$
x_{0}^{2}+y_{0}^{2}-2 r^{2}=0
$$

from which it is clear that the geometric locus of the intersection point $P$ of the perpendicular tangents is a circle concentric with the origin, whose radius is:

$$
R=r \sqrt{2}
$$

In our case, $r^{2}=32$, so the equation of the geometric locus from (5) is:

$$
x^{2}+y^{2}=64
$$

(Antal Márkus.)