Ans: B
Note that $A$ and $B$ are points lying on the circle. Therefore $AB$ is a chord of the circle.
Note that $A$ and $B$ are points lying on the circle. Therefore $AB$ is a chord of the circle.
Note also that $P$ be a moving point such that $AP=BP$. Therefore the locus of $P$ is the perpendicular bisector of $AB$.
Hence, the locus of $P$ is the perpendicular bisector of a chord of the circle. Therefore, the locus of $P$ must pass through the centre of the circle.
Note that the centre of the circle
$\begin{array}{cl}
= & \left( -\dfrac{-6}{2}, – \dfrac{-4}{2} \right) \\
= & (3, 2)
\end{array}$
Sub. $(3,2)$ into the equation of the locus of $P$, we have
$\begin{array}{rcl}
(3) + 2(2) + k & = & 0 \\
k & = & -7
\end{array}$
