Ans: C
Since the circle passes through the point $(-2,3)$, substitute $(-2,3)$ into the equation of the circle. Hence, we have
Since the circle passes through the point $(-2,3)$, substitute $(-2,3)$ into the equation of the circle. Hence, we have
$\begin{array}{rcl}
(-2)^2 + (3)^2 +a(-2) -6(3) -3 & = & 0 \\
-2a -8 & = & 0 \\
a & = & 4
\end{array}$
Therefore, the equation of the circle is $x^2 + y^2 +4x-6y-3=0$. Hence, the radius of the circle
$\begin{array}{cl}
= & \sqrt{(\dfrac{4}{2})^2 + (\dfrac{-6}{2})^2 – (-3)} \\
= & 4
\end{array}$
Therefore, the area of the circle
$\begin{array}{cl}
= & \pi (4)^2 \\
= & 16 \pi
\end{array}$
