If the circles $x^2+y^2+kx+4y+2=0\text{ and }2\left(x^2+y^2\right)-4x-3y+k=0$ cut orthogonally, then $k=$

Equations of given circles are $x^2+y^2+kx+4y+2=0,x^2+y^2-2x-\frac{3}{2}y+\frac{k}{2}=0$

The two circles cut orthogonally $\Rightarrow 2g{g}^{\mathrm{\prime }}+2f{f}^{\mathrm{\prime }}=c+c^{\mathrm{\prime }}\Rightarrow k(-1)+4\left(-\frac{3}{4}\right)=2+\frac{k}{2}$

$\Rightarrow -k-3=2+k/2\Rightarrow k+k/2=-5\Rightarrow 3k/2=-5\Rightarrow k=-10/3$