2008-II-53

$\begin{array}{cl}
= & \left( -\dfrac{-4}{2}, – \dfrac{-8}{2} \right) \\
= & (2, 4)
\end{array}$

II 為正確。該圓的半徑

$\begin{array}{cl}
= & \sqrt{(2)^2 + (4)^2 – 11} \\
= & \sqrt{9} \\
= & 3
\end{array}$

III 為正確。原點與圓心的距離

$\begin{array}{cl}
= & \sqrt{(2-0)^2 – (4-0)^2} \\
= & \sqrt{20} \\
> & 3
\end{array}$

所以，原點與圓心的距離大於半徑。由此，原點在該圓以外。