Question

Find the value of k, if the points $(4,2)$ and $(k,-3)$ are conjugate points with respect to the circle $x^2+y^2-5x+8y+6=0$.

Answer 1

Equation of circle-

$x^2+y^2-5x+8y+6=0$

Polar of point $(4,2)$ is-

$S_1=0$

$4x+2y-5\left(\frac{x+4}{2}\right)+8\left(\frac{y+2}{2}\right)+6=0$

$4x+2y-\frac{5}{2}x-10+4y+8+6=0$

$3x+12y+8=0.....\left(1\right)$

Given that the points $(4,2)$ and $(k,-3)$ are conjugaet.

Therefore, Polar of point $(4,2)$ passes through the point $(k,-3)$.

$\therefore 3k+12(-3)+8=0$

$\Rightarrow 3k=36-8$

$\Rightarrow k=\frac{28}{3}$

Hence the value of $k$ is $\frac{28}{3}$.