Question

If kx+y+1=0 & x-y+2=0 are conjugate lines w.r.to the circle x2+y2+4x-4y-1=0 then k=?

Answer 1

The two lines are

$\quad\quad kx+y+1=0\quad.....(1)$

$\quad\quad x-y+2=0\quad.....(2)$

Let $(x_{1},\:y_{1})$ be the pole of the second straight line. Then the equation of the polar of the point $(x_{1},\:y_{1})$ with respect to the circle $x^{2}+y^{2}+4x-4y-1=0$ is

$\quad xx_{1}+yy_{1}+2(x+x_{1})-2(y+y_{1})-1=0$

$\Rightarrow xx_{1}+yy_{1}+2x+2x_{1}-2y-2y_{1}-1=0$

$\Rightarrow (x_{1}+2)x+(y_{1}-2)y+(2x_{1}-2y_{1}-1)=0\quad.....(3)$

Here $(3)$ no. straight line is identical with the straight line $(2)$. Then

$\quad\frac{x_{1}+2}{1}=\frac{y_{1}-2}{-1}=\frac{2x_{1}-2y_{1}-1}{2}$

So, $\frac{x_{1}+2}{1}=\frac{2x_{1}-2y_{1}-1}{2}$

$\Rightarrow 2x_{1}+4=2x_{1}-2y_{1}-1$

$\Rightarrow 2y_{1}=-5$

$\Rightarrow \color{blue}y_{1}-\frac{5}{2}$

and $\frac{y_{1}-2}{-1}=\frac{2x_{1}-y_{1}-1}{2}$

$\Rightarrow -2y_{1}+4=2x_{1}-2y_{1}-1$

$\Rightarrow 2x_{1}=5$

$\Rightarrow \color{blue}x=\frac{5}{2}$

In order that two straight lines be conjugate, the pole $\left(\frac{5}{2},\:-\frac{5}{2}\right)$ of the second straight line lies on the first line.

Hence, $k\left(\frac{5}{2}\right)+\left(-\frac{5}{2}\right)+2=0$

$\Rightarrow \frac{5k}{2}+\frac{5}{2}+1=0$

$\Rightarrow \frac{5k}{2}=\frac{3}{2}$

$\Rightarrow \color{blue}{k=\frac{3}{5}}$

Answer: The value of $k$ is $\frac{3}{5}$.