If the lines $2 x+y+12=0, k x-3 y-10=0$ are conjugate with respect to the circle $x^2+y^2-4 x+3 y-1=0$, then…
If the lines $2 x+y+12=0, k x-3 y-10=0$ are conjugate with respect to the circle $x^2+y^2-4 x+3 y-1=0$, then $k=$
4
-9
-3
-5
Solution
Lines $2 x+y+12=0, k x-3 y-10=0$ are conjugate with respect to given circle.
$\therefore$ Pole of line $2 x+y+12=0$ lie on the line $k x-3 y-10=0$
Equation of polar of pole $(x, y)$ to circle $x^2+y^2-4 x+3 y-1=0$ is
$
\begin{array}{r}
x x_1+y y_1-2\left(x+x_1\right)+\frac{3}{2}\left(y+y_1\right)-1=0 \\
\Rightarrow x\left(x_1-2\right)+y\left(y_1+\frac{3}{2}\right)-2 x_1+\frac{3}{2} y_1-1=0
\end{array}
$
it is same as $2 x+y+12=0$
So, $\frac{x_1-2}{2}=\frac{y_1+\frac{3}{2}}{1}=\frac{-2 x_1+\frac{3}{2} y_1-1}{12}$
$
\Rightarrow \quad x_1-2=2 y_1+3 \text { and } 12 y_1+18=-2 x_1+\frac{3}{2} y_1-1
$
$
\begin{aligned}
\left(\frac{21}{2}+4\right) y_1 & =-29 \\
\left(\frac{29}{2}\right) y_1 & =-29 \Rightarrow y_1=-2
\end{aligned}
$
put in Eq. (i)
$
x_1+4=5 \Rightarrow x_1=1
$
Point $\left(x_1, y_1\right)$ lie of $k x-3 y-10=0$
$
\begin{array}{ll}
\Rightarrow & k(1)-3(-2)-10=0 \\
\Rightarrow & k+6-4=0 \Rightarrow k=4
\end{array}
$
