If $\mathrm{Re}\left(\frac{z-1}{2z+i}\right)=1$, where z = x + iy, then the point (x, y) lies on a

Let $z=x+iy$.  Then

$\left(\frac{z-1}{2z+i}\right)=\frac{\left(x-1\right)+iy}{2\left(x+iy\right)+i}=\frac{\left(x-1\right)+iy}{2x+\left(2y-1\right)i}\times \frac{2x-\left(2y+1\right)i}{2x-\left(2y+1\right)}$

Now $\mathrm{Re}\left(\frac{z+1}{2z+i}\right)=\frac{2x\left(x-1\right)+y\left(2y+1\right)}{{\left(2x\right)}^2+{\left(2y+1\right)}^2}=1$

$\Rightarrow 2x^2+2y^2-2x+y=4x^2+4y^2+4y+1\Rightarrow 2x^2+2y^2+2x+3y+1=0$

$\Rightarrow x^2+y^2+x+\frac{3}{2}y+\frac{1}{2}=0$

circle with centre $\left(-\frac{1}{2},-\frac{3}{4}\right)$

$r=\sqrt{\frac{1}{4}+\frac{9}{16}-\frac{1}{2}}=\sqrt{\frac{4+9-8}{16}}=\frac{\sqrt{5}}{4}$