Let $x^2+y^2+Dx+Ey+F=0$ be the equation of $C$. Since $C$ passes through $(0,0)$, $(10,-24)$ and $(17,-7)$, we have
$\left\{ \begin{array}{ll}
0^2+0^2+D(0)+E(0) +F =0 & \ldots \unicode{x2460} \\
10^2+(-24)^2 +D(10) +E(-24) +F = 0 & \ldots \unicode{x2461} \\
17^2 +(-7)^2 +D(17) +E(-7) +F =0 & \ldots \unicode{x2462}
\end{array}\right.$
From $\unicode{x2460}$, we have $F=0$.
From $\unicode{x2461}$, we have
$\begin{array}{rcl}
100+576 +10D-24E & = & 0 \\
5D -12E+338 & = & 0 \ldots \unicode{x2463}
\end{array}$
From $\unicode{x2462}$, we have
$\begin{array}{rcl}
289+49 +17D -7E & = & 0 \\
17D -7E+338 & = & 0 \ldots \unicode{x2464}
\end{array}$
$17 \times \unicode{x2463} -5\times \unicode{x2464}$, we have
$\begin{array}{rcl}
-169E +4056 & = & 0 \\
E & = & 24
\end{array}$
Sub. $E=24$ into $\unicode{x2463}$, we have
$\begin{array}{rcl}
5D -12(24) +338 & = & 0 \\
D & = & -10
\end{array}$
Therefore, the equation of $C$ is $x^2+y^2 -10x+24y=0$.
A is not true. Note that the coordinates of the centre and the radius of $C$ are $(5,-12)$ and $13$ respectively.
$\begin{array}{rcl}
PQ & = & \sqrt{(10-17)^2+(-24-(-7))^2} \\
PQ & = & \sqrt{338} \\
PQ & \neq & 26
\end{array}$
B is not true. The area of $C$
$\begin{array}{cl}
= & \pi (13)^2 \\
= & 169\pi
\end{array}$
C is true. Sub. $(16,-9)$ into the left side of the equation of $C$, we have
$\begin{array}{cl}
& 16^2 +(-9)^2 -10(16) +24(-9) \\
= & -39 \\
< & 0
\end{array}$
$\therefore (16,-9)$ lies inside $C$.
D is not true. Sub. $(5,-12)$ into the left side of $5x+12y=0$, we have
$\begin{array}{cl}
& 5(5) +12(-12) \\
= & -119 \\
\neq & 0
\end{array}$
$\therefore$ the centre of $C$ does not lie on $5x+12y=0$.
