Since, $P(-2,2)$ and $Q(k, 0)$ is the end points of a diameter.
$\therefore $ Mid-point $M$ of $P Q=\frac{-2+k}{2}, \frac{2+0}{2}$
i.e., $ M=\left(\frac{k}{2}-1,1\right)$
Also, $M P=$ radius of circle
$\therefore \sqrt{\left(-2-\frac{k}{2}+1\right)^{2}+(2-1)^{2}}=1$
$\Rightarrow \sqrt{\left(-1-\frac{k}{2}\right)^{2}+1}=1$
$\Rightarrow 1+\frac{k^{2}}{4}+k+1=1$
$\Rightarrow k^{2}+4 k+4=0$
$\Rightarrow (k+2)^{2}=0$
$\Rightarrow k=-2$
$\therefore $ Mid-point $M=\left(-\frac{2}{2}-1,1\right)=(-2,1)$
which is equal to the centre of circle.
$\therefore $ Equation of circle is
$(x+2)^{2}+(y-1)^{2}=1^{2}$
$\Rightarrow x^{2}+4 x+4+y^{2}-2 y+1=1$
$\Rightarrow x^{2}+y^{2}+4 x-2 y+4=0$