Ans: (a) $(x-2)^2+(y+1)^2=100$ (c) (i) $F$, $G$ and $H$ are collinear. (ii) $12x+5y-19=0$
- Note that the radius of $C$
$\begin{array}{cl}
= & \sqrt{(-6-2)^2 + (5-(-1))^2} \\
= & 10
\end{array}$Hence, the equation of $C$ is
$\begin{array}{rcl}
(x-2)^2 + (y-(-1))^2 & = & 10^2 \\
x^2 -4x +4 + y^2 +2y +1 – 100 & = & 0 \\
x^2 + y^2 -4x +2y – 95 & = & 0
\end{array}$ - The distance between $F$ and $G$
$\begin{array}{cl}
= & \sqrt{(-3-2)^2 + (11-(-1))^2} \\
= & 13 \\
> & 10 \text{, which is the radius of $C$}
\end{array}$Therefore, $F$ lies outside $C$.
-
- $F$, $G$ and $H$ are collinear.
- The required equation is
$\begin{array}{rcl}
\dfrac{y-(-1)}{x-2} & = & \dfrac{11-(-1)}{-3-2} \\
-5y-5 & = & 12x -24 \\
12x + 5y -19 & = & 0
\end{array}$
