The equations of the sides of a triangle are:

$$
y=x+1, \quad y=-\frac{1}{2} x-2, \quad y=3 x-9
$$

What is the equation of the circle passing through the vertices of the triangle?
We calculate the coordinates of the vertices of the triangle from the equations of its sides. By combining two equations at a time, we get:

$$
x_{1}=-2, y_{1}=-1, \quad x_{2}=5, \quad y_{2}=6, \quad x_{3}=2, y_{3}=-3
$$

Since the circle we are looking for passes through these points, the coordinates of these points satisfy the equation of the circle. Therefore, we substitute the coordinates of the vertices for $x$ and $y$ in the equation

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

Thus, we have:

$$
\begin{gathered}
(-2-a)^{2}+(-1-b)^{2}=r^{2} \\
(5-a)^{2}+(6-b)^{2}=r^{2} \\
(2-a)^{2}+(-3-b)^{2}=r^{2}
\end{gathered}
$$

From these equations, we get $a=2, b=2, r=5$, so the equation of the circle is:

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

(Guman Jenö, Losoncz.)