留意 $C$ 的一般式為 $x^2+y^2-2x+4y-\frac{5}{2}=0$。
I 為錯誤。$C$ 的半徑
$\begin{array}{cl}
= & \sqrt{(\dfrac{2}{2})^2 +(\dfrac{4}{2})^2-(-\dfrac{5}{2})} \\
= & \sqrt{\dfrac{15}{2}} \\
\neq & 5
\end{array}$
II 為正確。留意 $PQ$ 的中點
$\begin{array}{cl}
= & \left( \dfrac{-1+4}{2}, \dfrac{2+0}{2} \right) \\
= & (\dfrac{3}{2}, 1)
\end{array}$
把以上的坐標代入 $C$ 的方程的左方，可得
$\begin{array}{cl}
& (\dfrac{3}{2})^2 +(1)^2-2(\dfrac{3}{2}) +4(1)-\dfrac{5}{2} \\
= & \dfrac{7}{4} \\
> & 0
\end{array}$
所以，$PQ$ 的中點在 $C$ 之外。
III 為正確。留意
$\begin{array}{rcl}
G & = & \left(-\dfrac{-2}{2},-\dfrac{4}{2}\right) \\
& = & (1,-2)
\end{array}$
考慮 $\Delta PGQ$，
$\begin{array}{rcl}
PQ & = & \sqrt{(-1-4)^2+(2-0)^2} \\
& = & \sqrt{29} \\
PG & = & \sqrt{(-1-1)^2+(2-(-2))^2} \\
& = & \sqrt{20} \\
QG & = & \sqrt{(4-1)^2+(0-(-2))^2} \\
& = & \sqrt{13}
\end{array}$
則利用餘弦公式，可得
$\begin{array}{rcl}
\cos \angle PGQ & = & \dfrac{PG^2+QG^2-PQ^2}{2(PG)(QG)} \\
& = & \dfrac{20+13-29}{2(\sqrt{20})(\sqrt{13})} \\
& = & 0.124~034~734 \\
\angle PGQ & = & 82.874~983~65^\circ
\end{array}$
所以，$\angle PGQ$ 為一銳角。
