The equations of the sides of a certain triangle are:

$$
x-3 y=2, \quad 7 x-y=34, \quad x+2 y=-8
$$

Determine the equation of the circle circumscribed around the triangle without calculating the coordinates of the vertices.
I. Solution: We calculate the cosine of the angles of the triangle, then express the distances from the circumcircle center to each side in two different ways. This gives us three equations from which the two coordinates of the circumcircle center and the radius of the circle can be calculated.

Let the angles of the triangle be $\omega_{1}, \omega_{2}, \omega_{3}$. The absolute values of their tangents are derived from the direction factors $m_{1}=\frac{1}{3}, m_{2}=7, m_{3}=-\frac{1}{2}$:

$$
\left|\operatorname{tg} \omega_{1}\right|=\left|\frac{m_{2}-m_{3}}{1+m_{1} m_{2}}\right|=\frac{7+\frac{1}{2}}{1-\frac{7}{2}}=3, \text { similarly }\left|\operatorname{tg} \omega_{2}\right|=1,\left|\operatorname{tg} \omega_{3}\right|=2
$$

Since the absolute value of each tangent is at least 1, the triangle is acute-angled, and thus the cosine of each angle is positive.

Based on $\cos \omega=\frac{1}{\sqrt{1+\operatorname{tg}^{2} \omega}}$:

$$
\cos \omega_{1}=\frac{1}{\sqrt{10}}, \quad \cos \omega_{2}=\frac{1}{\sqrt{2}}, \quad \cos \omega_{3}=\frac{1}{\sqrt{5}}
$$

The distance from the circumcircle center to the sides of the triangle:

$$
d_{1}=r \cos \omega_{1}=\frac{r}{\sqrt{10}}, \quad d_{2}=\frac{r}{\sqrt{2}}, \quad d_{3}=\frac{r}{\sqrt{5}}
$$

Let the coordinates of the center be $x_{0}$ and $y_{0}$. The distance from the circumcircle center to the sides:

$$
d_{1}=\frac{x_{0}-3 y_{0}-2}{\sqrt{10}}, \quad d_{2}=\frac{y_{0}-7 x_{0}+34}{5 \sqrt{2}}, \quad d_{3}=\frac{x_{0}+2 y_{0}+8}{\sqrt{5}}
$$

By comparing equations (1) and (2):

$$
r=x_{0}-3 y_{0}-2, \quad 5 r=y_{0}-7 x+34, \quad r=x_{0}+2 y_{0}+8
$$

Solving the resulting system of equations:

$$
x_{0}=1, \quad y_{0}=-2, \quad r=5
$$

Thus, the equation of the circumcircle of the triangle is:

$$
(x-1)^{2}+(y+2)^{2}=25
$$

Pál Bártfai (Bp., I., Petőfi g. III. o. t.)

II. Solution: Let the equations of the three lines be $L_{1}=0, L_{2}=0, L_{3}=0$.

Form the following expression:

$$
K=L_{1} L_{2}+\lambda L_{2} L_{3}+\mu L_{3} L_{1}
$$

where $\lambda$ and $\mu$ represent arbitrary real values.

The expression written is quadratic (the product of two linear expressions). For any chosen $\lambda$ and $\mu$, $K=0$ is the equation of a conic section. Different values of $\lambda$ and $\mu$ represent different conic sections.

The conic section represented by the equation $K=0$ passes through the intersection points of the given lines. For example, substituting the coordinates of the intersection point of the lines $L_{1}=0$ and $L_{2}=0$, regardless of the values of $\lambda$ and $\mu$, the expression $K$ will not only have its first term equal to 0, but also its second term (since $L_{2}=0$) and its third term (since $L_{1}=0$). Our task is to choose $\lambda$ and $\mu$ such that $K=0$ is precisely the equation of a circle. The condition for this is that the coefficients of the $x^{2}$ and $y^{2}$ terms be equal, and that the coefficient of the $x y$ term be 0.

Write out $K$ in detail.

$$
\begin{gathered}
K=(x-3 y-2)(7 x-y+34)+\lambda(7 x-y-34)(x+2 y+8)+ \\
+\mu(x+2 y+8)(x-3 y-2)=0
\end{gathered}
$$

Rearranging:

$$
\begin{gathered}
(7+7 \lambda+\mu) x^{2}+(-22+13 \lambda-\mu) x y+(3-2 \lambda-6 \mu) y^{2}+ \\
+(-48+22 \lambda+6 \mu) x+(104-76 \lambda-28 \mu) y+(68-272 \lambda-16 \mu)=0
\end{gathered}
$$

According to our condition:

$$
\begin{gathered}
7+7 \lambda+\mu=3-2 \lambda-6 \mu \\
\quad-22+13 \lambda-\mu=0
\end{gathered}
$$

From these equations:

$$
\lambda=\frac{3}{2}, \quad \mu=-\frac{5}{2}
$$

Substituting the obtained values into $K$, after rearrangement, we get:

$$
15 x^{2}+15 y^{2}-30 x+60 y-300=0
$$

Simplifying by 15:

$$
(x-1)^{2}+(y+2)^{2}=25
$$

It is possible that the equations used to determine the parameters are not independent, making them unsuitable for calculating $\lambda$ and $\mu$. This happens if two of the given lines are parallel, in which case the problem has no solution. (In this case, one vertex of the triangle is at infinity, and the circumcircle degenerates into a finite line (a triangle side) and the plane's line at infinity.)

Károly Bauer (Pécs, Bányaip. techn. IV. o. t.)