Ans: B
Rewrite the equation of the circle to general form, we have $x^2+y^2+4x-6y+\dfrac{3}{2}=0$.
Rewrite the equation of the circle to general form, we have $x^2+y^2+4x-6y+\dfrac{3}{2}=0$.
I is true. The centre of the circle
$\begin{array}{cl}
= & (-\dfrac{4}{2},-\dfrac{-6}{2}) \\
= & (-2,3)
\end{array}$
II is false. The radius of the circle
$\begin{array}{cl}
= & \sqrt{(-2)^2+(3)^2-4(\dfrac{3}{2})} \\
= & \sqrt{\dfrac{23}{2}}
\end{array}$
III is true. The distance between $(2,3)$ and the centre
$\begin{array}{cl}
= & \sqrt{(2-(-2))^2+(3-3)^2} \\
= & 4 \\
> & \sqrt{\dfrac{23}{2}}\mbox{, which is the radius of the circle.}
\end{array}$
