8,9 |
| $[\underline{\text { Circle (other) }}]$ |  |

Form the equation of the circle passing through the points $A(-2 ; 1), B(9 ; 3)$ and $C(1 ; 7)$.
Since

$$
\overrightarrow{A C}=\overrightarrow{(1-(-2) ; 7-1)}=\overrightarrow{(3 ; 6)}, \overrightarrow{B C}=\overrightarrow{(1-9) ; 7-3)}=\overrightarrow{(-8 ; 4)}
$$

then

$$
\overrightarrow{A C} \cdot \overrightarrow{B C}=3 \cdot(-8)+6 \cdot 4=0
$$

Therefore, $A C \perp B C$, i.e., triangle $A B C$ is a right triangle. Thus, the center of its circumscribed circle coincides with the midpoint $M\left(x_{0} ; y_{0}\right)$ of the hypotenuse $A B$, and the radius $R$ of this circle is half the length of segment $A B$.

Since

$$
\begin{gathered}
x_{0}=\frac{-2+9}{2}=\frac{7}{2}, y_{0}=\frac{1+3}{2}=2 \text {, and } \\
R=\frac{1}{2} A B=\frac{1}{2} \sqrt{(9-(-2))^{2}+(3-1)^{2}}=\frac{1}{2} \sqrt{121+4}=\frac{\sqrt{125}}{2}
\end{gathered}
$$

the desired equation is

$$
\left(x-\frac{7}{2}\right)^{2}+(y-2)^{2}=\frac{125}{4}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_198acf192511df7a04ddg-12.jpg?height=572&width=686&top_left_y=2029&top_left_x=686)

## Answer

$\left(x-\frac{7}{2}\right)^{2}+(y-2)^{2}=\frac{125}{4}$.

[ Area of a triangle (using two sides and the angle between them). ]

Problem