Ans: C
Note that the equation of $C$ in general form is $x^2 + y^2 -4x+10y+\dfrac{65}{3}=0$.
Note that the equation of $C$ in general form is $x^2 + y^2 -4x+10y+\dfrac{65}{3}=0$.
I is not true. The radius of $C$
$\begin{array}{cl}
= & \sqrt{\left(\dfrac{-4}{2}\right)^2 + \left(\dfrac{10}{2} \right)^2 – \dfrac{65}{3}} \\
= & \sqrt{\dfrac{22}{3}}
\end{array}$
II is true. Substitute $(0,0)$ into the left hand side of the equation of $C$, we have
$\begin{array}{cl}
& 3(0)^2 + 3(0)^2 -12(0) + 30(0) + 65 \\
= & 65 \\
> & 0
\end{array}$
Therefore, the origin lies outside of $C$.
III is true. The coordinates of the centre of $C$
$\begin{array}{cl}
= & \left( -\dfrac{-4}{2}, – \dfrac{10}{2} \right) \\
= & (2,-5)
\end{array}$
