Real part of $\frac{\bar{z}+2}{\bar{z}-1}$ is given by
$\frac{1}{2}\left[\frac{\bar{z}+2}{\bar{z}-1}+\left(\frac{\bar{z}+2}{\bar{z}-1}\right)\right]=4$
$\Rightarrow \frac{\bar{z}+2}{\bar{z}-1}+\frac{z+2}{z-1}=8$
$\Rightarrow z \bar{z}-\bar{z}+2 z-2+z \bar{z}+2 \bar{z}-z-2$
$=8(z \bar{z}-\bar{z}-z+1)$
$\Rightarrow z \bar{z}-\frac{3}{2} z-\frac{3}{2} \bar{z}+2=0)$ ...(i)
Comparing with the equation
$z \bar{z}+\bar{a} z+ a \bar{z}+b=0$
we get $a=-\frac{3}{2}$ and $b =2$.
Thus, the locus of $z$ given by the equation (i) is a circle with center
$\frac{3}{2}$ and radius $=\frac{1}{2}$