Question

If the lines $2x+y+12=0,kx-3y-10=0$ are conjugate with respect to the circle $x^2+y^2-4x+3y-1=0$, then $k=$

Answer 1

Let pole be $(m,n)$

$T=0$

$xm+yn-4\left(\frac{x+m}{2}\right)+3\left(\frac{y+n}{2}\right)-1=0$

$mx+ny-2x-2m+\frac{3y}{2}+\frac{3n}{2}-1=0$

$(m-2)x+(n+\frac{3}{2})y+\frac{3n}{2}y+\frac{3n}{2}-2m-1=0$

$2x+y+12=0$

$\frac{m-2}{2}=\frac{n+\frac{3}{2}}{1}=\frac{3\frac{n}{2}-2m-1}{12}$

$m-2=2n+3$ $\frac{m-2}{2}=\frac{\frac{3n}{2}-2m-1}{12}$

$m-2n=5$ ....(1)$\times 16$ $6m-12=\frac{3n}{2}-2m-1$

$16m-3n=22$ .....(2) $12m-24=3n-4m-2$

$16m-32n=80$

$16m-3n=22$

$-$ $+$ $-$

$\overline{-29n=58}$

$n=-2$

$m=5+2n=5-4=1$

$m=1$

$(m,n)$ lies on $kx-3y-10-0$

$k-3(-2)-10=0\Rightarrow k+6-10=0$

$k=4$.