The equation of an ellipse is given: $25 x^{2}+36 y^{2}=900$. At point $M$ on this ellipse $\left(x_{1}=4.8, y_{1}>0\right)$, we draw a tangent, which intersects the extended minor axis at point $N$. A circle is drawn through points $M$ and $N$, with its center on the minor axis. Determine the coordinates of the points where the circle intersects the major axis of the ellipse.
Substituting the value of $x_{1}$ into the equation of the ellipse, we get

$$
25 \cdot 4.8^{2}+36 y_{1}^{2}=900
$$

from which the positive value of $y_{1}$ is: $y_{1}=3$. Therefore, the equation of the tangent line drawn at point $M$ is:

$$
25 \cdot 4.8 x+36 \cdot 3 y=900
$$

or

$$
120 x+108 y=900
$$

Let's find the coordinates of the intersection point of this line with the ordinate axis. Since the abscissa of this point is $x_{2}=0$, then

$$
108 y_{2}=900
$$

from which

$$
y_{2}=\frac{25}{3}
$$

Thus, the coordinates of point $N$ are: $x_{2}=0, y_{2}=\frac{25}{3}$. The center of the circle passing through points $M$ and $N$ is on the ordinate axis; hence its abscissa is 0; if its ordinate is $q$, then the equation of this circle is:

$$
x^{2}+(y-q)^{2}=r^{2}
$$

To determine $q$ and $r$, substitute the coordinates of points $M$ and $N$ for $x$ and $y$; then

$$
4.8^{2}+(3-q)^{2}=r^{2}
$$

and

$$
\left(\frac{25}{3}-q\right)^{2}=r^{2}
$$

from which $q=\frac{263}{75}$ and $r=\frac{362}{75}$. Substituting these values into (1):

$$
x^{2}+\left(y-\frac{263}{75}\right)^{2}=\left(\frac{362}{75}\right)^{2}
$$

Let's finally find the coordinates of the points where this circle intersects the abscissa axis; since for these points $y=0$, then

$$
\begin{gathered}
x= \pm \sqrt{\left(\frac{362}{75}\right)^{2}-\left(\frac{263}{75}\right)^{2}}= \pm \sqrt{\frac{362+263}{75} \cdot \frac{362-263}{75}}= \\
= \pm \sqrt{11}= \pm 3.31 \ldots
\end{gathered}
$$

(János Prohászka, Prague.)

The problem was also solved by: Krausz B., Lukhaub Gy.