I is true. Rewrite the equation of $C_1$, then the equation of $C_1$ becomes $x^2+y^2+2x+4y-\dfrac{149}{2} = 0$.
The centre of $C_1$
$\begin{array}{cl}
= & \left(\dfrac{-2}{2}, \dfrac{-4}{2}\right) \\
= & (-1,-2)
\end{array}$
Sub. the centre of $C_1$ into the left side of the equation of $C_2$, we have
$\begin{array}{cl}
& (-1)^2 + (-2)^2 -8(-1) – 20(-2) -53 \\
= & 0
\end{array}$
Therefore, the centre of $C_1$ lies on $C_2$.
II is not true. The radius of $C_1$
$\begin{array}{cl}
= & \sqrt{(-1)^2 + (-2)^2 – (-\dfrac{149}{2})} \\
= & \sqrt{\dfrac{159}{2}}
\end{array}$
The centre of $C_2$
$\begin{array}{cl}
= & \left(-\dfrac{-8}{2}, -\dfrac{-20}{2} \right) \\
= & (4, 10)
\end{array}$
The radius of $C_2$
$\begin{array}{cl}
= & \sqrt{(4)^2 + (10)^2 – (-53)} \\
= & 13
\end{array}$
Therefore, the radii of $C_1$ and $C_2$ are not equal.
III is true. The distance between two centres
$\begin{array}{cl}
= & \sqrt{(-1 -4)^2 +(-2-10)^2} \\
= & 13 \\
\end{array}$
The sum of the radii of the circles
$\begin{array}{cl}
= & 13 + \sqrt{\dfrac{159}{2}} \\
= & 21.916\ 277\ 25
> & 13
\end{array}$
Hence, $C_1$ and $C_2$ intersect at two distinct points.
