Equation of circle is $x^2+y^2-2x+4y-2=0$

Given $P\left(x_1,y_1\right)\equiv \left(2,3\right)$ & $Q\left(K,-2\right)$ are conjugate with respect to given circle.

Equation of polar of $P$ is $x{x}_1+y{y}_1-2\left(\frac{x+x_1}{2}\right)+4\left(\frac{y+3}{2}\right)-2=0$

$\Rightarrow 2x+3y-1\left(x+2\right)+2\left(y+3\right)-2=0$

$\Rightarrow 2x+3y-x-2+2y+6-2=0$

$\Rightarrow x+5y+2=0$

$\because P\left(2,3\right) \& Q\left(K,-2\right)$ are conjugate, so polar of point $P$ passes through $Q$.

$\therefore K-10+2=0$

$\therefore K=8$