A is not true. Sub. $(0, 0)$ into the left side of the equation of $C$, we have
$\begin{array}{rcl}
\text{LS} & = & 5(0)^2 + 5(0)^2 -30(0) + 10(0) +6 \\
& = & 6 \\
& > & 0
\end{array}$
Therefore, the origin lies outside $C$.
B is not true. Rewrite the equation of the circle $C$ to general form, we have $x^2 +y^2 -6x + 2y + \dfrac{6}{5} = 0$.
Note that the centre of $C$
$\begin{array}{cl}
= & \left(-\dfrac{-6}{2}, -\dfrac{2}{2} \right) \\
= & (3,-1)
\end{array}$
and the radius of $C$
$\begin{array}{cl}
= & \sqrt{(3)^2 + (-1)^2 – \dfrac{6}{5}} \\
= & \sqrt{\dfrac{44}{5}}
\end{array}$
Therefore, $C$ lies in the first and the forth quadrant.
C is true. The circumference of $C$
$\begin{array}{cl}
= & 2 \pi \sqrt{\dfrac{44}{5}} \\
= & 18.638\ 939\ 75 \\
< & 20
\end{array}$
D is not true. The coordinates of the centre of $C$ are $(3,-1)$.
