There is a circle and a line:

$$
x^{2}+y^{2}-10 x-12 y+45=0, \quad y=3 x
$$

Determine the equation of the tangent lines to the circle that are parallel to the given line.
I. solution: The equation of the circle can also be written as:

$$
(x-5)^{2}+(y-6)^{2}=16
$$

from which the center of the circle is $O(5,6)$ and the radius is $r=4$.

The line passing through the center of the circle and perpendicular to the given line $y=3 x$ is:

$$
y-6=-\frac{1}{3}(x-5)
$$

which intersects the circumference of the circle at the points of tangency.

From (1),

$$
x=23-3 y
$$

substituting this value into the equation of the circle,

$$
(18-3 y)^{2}+(y-6)^{2}=16
$$

or

$$
10 y^{2}-120 y+344=0
$$

from which

$$
y_{1,2}=6 \pm \frac{2}{5} \sqrt{10}
$$

and thus

$$
x_{1,2}=23-18 \mp \frac{6}{5} \sqrt{10}=5 \mp \frac{6}{5} \sqrt{10}
$$

Therefore, the points of tangency are:

$$
E_{1}\left(5-\frac{6}{5} \sqrt{10}, 6+\frac{2}{5} \sqrt{10}\right) \quad \text { and } \quad E_{2}\left(5+\frac{6}{5} \sqrt{10}, 6-\frac{2}{5} \sqrt{10}\right)
$$

The equation of the line passing through $E_{1}$ and parallel to the given line is:

$$
y-6-\frac{2}{5} \sqrt{10}=3\left(x-5+\frac{6}{5} \sqrt{10}\right)
$$

from which

$$
y=3 x-9+4 \sqrt{10}
$$

Similarly, we obtain the tangent passing through $E_{2}$:

$$
y=3 x-9-4 \sqrt{10}
$$

Klofszky Emil (Győr, Révai g. III. o. t.)

II. solution: Let the equation of the tangent be $y=m x+b$. Because of the parallelism, $m=3$. The line $y=3 x+b$ touches the circle if it intersects at two coinciding points, that is, if substituting the value of $y$ into the equation of the circle:

$$
x^{2}+(3 x+b)^{2}-10 x-12(3 x+b)+45=0
$$

the discriminant of the resulting quadratic equation in $x$ is 0.

Rearranging the equation under (1),

$$
10 x^{2}+(6 b-46) x+\left(b^{2}-12 b+45\right)=0
$$

The discriminant is:

$$
(6 b-46)^{2}-40\left(b^{2}-12 b+45\right)=0
$$

or

$$
b^{2}+18 b-79=0
$$

from which

$$
b_{1,2}=-9+4 \sqrt{10}
$$

Therefore, the equations of the tangents are:

$$
\begin{aligned}
& y=3 x-9+4 \sqrt{10} \\
& y=3 x-9-4 \sqrt{10}
\end{aligned}
$$

III. solution: The normal form of the line $y=3 x+b$ is

$$
\frac{3 x-y+b}{ \pm \sqrt{10}}=0
$$

$b$ must be chosen so that the distance from this line to the point $O(5,6)$ is 4, that is,

$$
\frac{15-6+b}{ \pm \sqrt{10}}=4
$$

from which

$$
b_{1,2}=-9 \pm 4 \sqrt{10}
$$

and thus the equations of the tangents are:

$$
y=3 x-9+4 \sqrt{10}
$$

and

$$
y=3 x-9-4 \sqrt{10}
$$

Rockenbauer Magda (Bp. X., I. László g. III. o. t.)