Given the circle $O$ and the line $e$, on which the point $A$ moves. The polar of point $A$ with respect to the circle intersects the perpendicular line to $e$ at point $A$ at point $M$. What is the geometric locus of point $M$?
Let the origin of our coordinate system be the center $O$ of the circle; the $X$-axis should be perpendicular to $e$. Thus, the equations of the circle and the line are:

$$
\begin{gathered}
x^{2}+y^{2}=r^{2} \ldots \\
x=a \ldots
\end{gathered}
$$

where $a$ is the distance of the point $O$ from the line $e$.

The polar of a point $(\alpha, \beta)$ with respect to the circle 1) is given by the equation:

$$
\alpha x+\beta y-r^{2}=0
$$

In the given case, for point $A$, $\alpha=a$ is constant, only $\beta$ varies; the equation of the polar of $A$ is:

$$
a x+\beta y-r^{2}=0 \ldots
$$

The equation of the perpendicular to $e$ at point $A$ is:

$$
y=\beta \ldots
$$

The coordinates of the intersection point $M$ of the lines 3) and 4) satisfy the equations 3) and 4); if we eliminate the variable $\beta$ from these equations, we obtain the equation of the geometric locus of point $M$:

$$
a x+y^{2}-r^{2}=0 \quad \text { or } \quad y^{2}=r^{2}-a x \ldots
$$

This is the equation of a parabola whose focal axis is the $X$-axis (in the negative direction). Its vertex is at the abscissa $x_{0}=\frac{r^{2}}{a}$. This vertex $S$ is none other than the pole of the line $e$ with respect to the circle.

The parabola passes through the points $(x=0, y= \pm r)$, which are the points where the circle intersects the $Y$-axis.

The focus of the parabola is $\frac{a}{4}$ units away from the vertex (towards the center of the circle).

Since the polar of point $A$ passes through the point $S$ and is perpendicular to $O A$, the point $M$ corresponding to any point $A$ can be easily constructed once the point $S$ is determined.

Jenő Krisztonosich (Szent László g. VIII. o. Bp. X.)