Ans: B
I is not true. Rewrite the equation of $C$ to general form, we have $x^2 + y^2 +2x -6y + \dfrac{15}{2} = 0$. Then the radius
I is not true. Rewrite the equation of $C$ to general form, we have $x^2 + y^2 +2x -6y + \dfrac{15}{2} = 0$. Then the radius
$\begin{array}{cl}
= & \sqrt{(\dfrac{-2}{2})^2 + (\dfrac{-(-6)}{2})^2 – \dfrac{15}{2}} \\
= & \sqrt{\dfrac{5}{2}}
\end{array}$
Therefore, the area of $C$
$\begin{array}{cl}
= & \pi \times (\sqrt{\dfrac{5}{2}})^2 \\
= & 2.5 \pi
\end{array}$
II is true. Sub. $(-3, 3)$ into the left side of the equation of $C$, we have
$\begin{array}{rcl}
\text{LS} & = & (-3)^2 + 3^2 + 2(-3) – 6(3) + \dfrac{15}{2} \\
& = & \dfrac{3}{2} \\
& > & 0
\end{array}$
Therefore, the point $(-3, 3)$ lies outside $C$.
III is not true. The centre of $C$
$\begin{array}{cl}
= & \left( \dfrac{-2}{2}, \dfrac{-(-6)}{2} \right) \\
= & (-1, 3)
\end{array}$
Therefore, the centre of $C$ lies in quadrant II.
