- $\Gamma$ is the perpendicular bisector of $AB$.
-
- Since $\Gamma \perp AB$, then we have
$\begin{array}{rcl}
m_\Gamma \times m_{AB} & = & -1 \\
(-3) \times m_{AB} & = & -1 \\
m_{AB} & = & \dfrac{1}{3}
\end{array}$Therefore, the equation of the straight lines which passes through $A$ and $B$ is
$\begin{array}{rcl}
\dfrac{y-(-4)}{x-2} & = & \dfrac{1}{3} \\
3(y+4) & = & x-2 \\
3y +12 & = & x-2 \\
x-3y-14& = & 0
\end{array}$ - Note that the intersection of $\Gamma$ and $AB$ is the centre of the required circle.
$\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}$, we have
$\begin{array}{rcl}
10x-50 & = & 0 \\
10x & = & 50 \\
x & = & 5
\end{array}$Sub. $x=5$ into $\unicode{x2460}$, we have
$\begin{array}{rcl}
5-3y-14 & = & 0 \\
-3y & = & 9 \\
y & = & -3
\end{array}$Therefore, the coordinates of the centre of the required circle are $(5,-3)$.
The radius of the required circle
$\begin{array}{cl}
= & \sqrt{(5-2)^2 +(-3-(-4))^2} \\
= & \sqrt{10}
\end{array}$Hence, the equation of the required circle is
$\begin{array}{rcl}
(x-5)^2 +(y-(-4))^2 & = & (\sqrt{10})^2 \\
(x-5)^2 +(y+4)^2 & = & 10
\end{array}$
- Since $\Gamma \perp AB$, then we have
2023-I-10
Ans: (a) $\Gamma$ is the perpendicular bisector of $AB$. (b) (i) $x-3y-14=0$ (ii) $(x-5)^2+(y+3)^2=10$
