We draw a circle through the foci of an ellipse, which touches an arbitrary tangent of the ellipse at point $M$. What is the geometric locus of this $M$ point if the tangent continuously changes its position?
Let the endpoints of the major axis of the ellipse be denoted by $A$ and $A^{\prime}$, and let $P$ be a point on the ellipse where we draw a tangent. Let the feet of the perpendiculars from the foci of the ellipse $\left(F_{1}\right.$ and $\left.F_{2}\right)$ to the tangent be $G_{1}$ and $G_{2}$. It can be easily shown that the quadrilateral $A G_{1} G_{2} A^{\prime}$ is inscribed in a circle that is concentric with the ellipse and has a radius equal to the semi-major axis.

Indeed, $C G_{1} \| F_{2} F_{1}^{\prime}$, where $C$ is the center of the ellipse and $F_{1}^{\prime}$ is the intersection of the lines $F_{1} G_{1}$ and $F_{2} P$. From the known properties of the ellipse, $F_{2} F_{1}^{\prime} = 2a$, so $C G_{1} = a$. Similarly, it can be shown that $C G_{2} = a$.

Let the intersection of the lines $A A^{\prime}$ and $G_{1} G_{2}$ be $O$. Then, by the known properties of circles and secants,

$$
O G_{1} \cdot O G_{2} = O A \cdot O A^{\prime}
$$

or if we denote $O C$ by $d$:

$$
O G_{1} \cdot O G_{2} = (d - a)(d + a) = d^2 - a^2
$$

Furthermore,

$$
O G_{1} : O G_{2} = O F_{1} : O F_{2} = (d - c) : (d + c)
$$

If we now draw a circle that passes through $F_{1}$ and $F_{2}$ and is tangent to $O P$ at $M$, we can write for this circle:

$$
O M^2 = O F_{1} \cdot O F_{2} = (d - c)(d + c)
$$

Now let's determine the coordinates of the point $M$ using $C K$ and $K M$.

$$
K M^2 = O M^2 - O K^2
$$

and $O K$ is, by the similarity of triangles $O M K$ and $O F_{1} G_{1}$,

$$
O K = O M \frac{O G_{1}}{O F_{1}}
$$

so

$$
O K^2 = O M^2 \frac{O G_{1}^2}{O F_{1}^2}
$$

thus

$$
K M^2 = O M^2 \left(1 - \frac{O G_{1}^2}{O F_{1}^2}\right)
$$

but $O G_{1}^2$ from the equations 1) and 2) is

$$
O G_{1}^2 = \left(d^2 - a^2\right) \frac{d - c}{d + c}
$$

therefore

$$
K M^2 = O M^2 \left(1 - \frac{d^2 - a^2}{d^2 - c^2}\right) = \left(d^2 - c^2\right) - \left(d^2 - a^2\right) = a^2 - c^2 = b^2
$$

so

$$
K M = \pm b
$$

The geometric locus of the point $M$ is thus two lines, specifically the two tangents to the ellipse at the points $B$ and $B^{\prime}$.

(Meitner Elemér, fr. VIII. o. t. Budapest).

The problem was also solved by Sramkó Loránd, fg. VIII. o. t. Rimaszombat.