2023-I-10

$\begin{array}{rcl}
m_\Gamma \times m_{AB} & = & -1 \\
(-3) \times m_{AB} & = & -1 \\
m_{AB} & = & \dfrac{1}{3}
\end{array}$

所以，通過 $A$ 及 $B$ 的直線的方程為

$\begin{array}{rcl}
\dfrac{y-(-4)}{x-2} & = & \dfrac{1}{3} \\
3(y+4) & = & x-2 \\
3y +12 & = & x-2 \\
x-3y-14& = & 0
\end{array}$

$\left\{ \begin{array}{ll}
x-3y-14=0 & \ldots \unicode{x2460} \\
3x+y-12=0 & \ldots \unicode{x2461}
\end{array}\right.$

$\unicode{x2460} +3\times \unicode{x2461}$，可得

$\begin{array}{rcl}
10x-50 & = & 0 \\
10x & = & 50 \\
x & = & 5
\end{array}$

把 $x=5$ 代入 $\unicode{x2460}$，可得

$\begin{array}{rcl}
5-3y-14 & = & 0 \\
-3y & = & 9 \\
y & = & -3
\end{array}$

所以，所求的圓的圓心為 $(5,-3)$。

所求的圓的半徑

$\begin{array}{cl}
= & \sqrt{(5-2)^2 +(-3-(-4))^2} \\
= & \sqrt{10}
\end{array}$

由此，所求的圓的方程為

$\begin{array}{rcl}
(x-5)^2 +(y-(-4))^2 & = & (\sqrt{10})^2 \\
(x-5)^2 +(y+4)^2 & = & 10
\end{array}$