If $(6,-k)$ and $(-3,2)$ are conjugate points with respect to circle $x^2+y^2+6 x+4 y+12=0$, then $k$ equals
If $(6,-k)$ and $(-3,2)$ are conjugate points with respect to circle $x^2+y^2+6 x+4 y+12=0$, then $k$ equals
$\frac{-7}{4}$
$\frac{7}{4}$
$\frac{4}{7}$
$\frac{-4}{7}$
Solution
Points $(6,-k)$ and $(-3,2)$ are conjugate points to the circle $x^2+y^2+6 x+4 y+12=0$
$\therefore$ We know that if $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$ are conjugate points of circle
$
x^2+y^2+2 g x+2 f y+c=0
$
Then, $x_1 x_2+y_1 y_2+g\left(x_1+x_2\right)+f\left(y_1+y_2\right)+c=0$
$\therefore \quad(6)(-3)+(-k)(2)+3(6-3)+2(-k+2)+12=0$
$
-18-2 k+9-2 k+4+12=0
$
$\Rightarrow \quad 4 k=7 \Rightarrow k=7 / 4$