The equations of the sides of a triangle are

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

Determine the equation of the circumscribed circle.
I. solution: The intersection points of the sides: $A(2,-3), B(5,6), C(-2,-1)$.

The coordinates of the midpoint of side $BC$ are: $\frac{5-2}{2}, \frac{6-1}{2}$, so the midpoint is: $A_{1}\left(\frac{3}{2}, \frac{5}{2}\right)$. Similarly, the midpoint of side $AC$ is: $B_{1}(0,-2)$.

The equation of the line perpendicular to side $BC$ at point $A_{1}$ is

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

or

$$
y=-x+4
$$

The equation of the line perpendicular to side $AC$ at point $B_{1}$ is

$$
y+2=2(x-0)
$$

or

$$
y=2 x-2
$$

From (1) and (2), the intersection point of the two perpendicular bisectors, which is the center of the circumscribed circle, is: $K(2,2)$.

The radius of the circle can be found by calculating the distance from the center to any vertex of the triangle, for example, from $B(5,6)$:

$$
r^{2}=(5-2)^{2}+(6-2)^{2}=25
$$

Thus, the equation of the circle is

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

Endre Csáki (Győr, Révai g. III. o. t.)

II. solution: Let the equation of the desired circle be

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

The points $A, B, C$ lie on the circle, so their coordinates satisfy the equation of the circle:

$$
\begin{aligned}
& (2-u)^{2}+(-3-v)^{2}=r^{2} \\
& (5-u)^{2}+(6-v)^{2}=r^{2} \\
& (-2-u)^{2}+(-1-v)^{2}=r^{2}
\end{aligned}
$$

From (1) and (2),

$$
u+3 v=8
$$

From (2) and (3),

$$
u+v=4
$$

From these two equations,

$$
u=2 \quad \text { and } \quad v=2 \text {. }
$$

Substituting these values into (1),

$$
r^{2}=(2-2)^{2}+(-3-2)^{2}=25
$$

Thus, the equation of the circle is:

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

Eligius Schmidt (Bp. I., Fürst S. g. II. o. t.)

III. solution: Let the equation of the circle be

$$
x^{2}+y^{2}+a x+b y+c=0
$$

Substituting the coordinates of points $A, B, C$:

$$
\begin{gathered}
4+9+2 a-3 b+c=0 \\
25+36+5 a+6 b+c=0 \\
4+1-2 a-b+c=0
\end{gathered}
$$

The solution to this system of equations is:

$$
a=b=-4 . \quad c=-17
$$

Thus, the equation of the circle is

$$
x^{2}+y^{2}-4 x-4 y-17=0
$$

which agrees with the previous result.