Ans: (a) $4x-3y-24=0$ (b) Yes
- Let $P=(x,y)$.
$\begin{array}{rcl}
PA & = & PB \\
\sqrt{(5-x)^2 + (7-y)^2} & = & \sqrt{(13-x)^2 + (1-y)^2} \\
25 – 10x +x^2 + 49 -14y +y^2 & = & 169 -26x +x^2 +1 -2y + y^2 \\
16x -12y -96 & = & 0 \\
4x – 3y -24 & = & 0
\end{array}$Therefore, the equation of $\Gamma$ is $4x – 3y -24 =0$.
- Note that $H=(6,0)$ and $K=(0,8)$. Note also that $\angle HOK = 90^\circ$. Therefore by the converse of angle in semi circle, $HK$ is a diameter of the circle $C$. Hence, the radius
$\begin{array}{cl}
= & \dfrac{1}{2} \times \sqrt{(6-0)^2 + (0-8)^2} \\
= & 5
\end{array}$Therefore, the circumference of $C$
$\begin{array}{cl}
= & 2 \pi (5) \\
= & 10 \pi \\
= & 31.4 \\
> & 30
\end{array}$Hence, the claim is correct.
