If R e z - 1 2 z + i = 1 , where z = x + i y , then the point x , y lies on a

Let $z=x+iy$
$\left(\frac{z-1}{2z+i}\right)=\frac{\left(x-1\right)+iy}{2\left(x+iy\right)+i}=\frac{\left(x-1\right)+iy}{2x+\left(2y+1\right)i}\times \frac{2x-\left(2y+1\right)i}{2x-\left(2y+1\right)i}$
$Re\left(\frac{z+1}{2z+i}\right)=\frac{2x\left(x-1\right)+y\left(2x+1\right)}{\left(2x^2\right)+{\left(2y+1\right)}^2}=1$
$\Rightarrow 2x^2+2y^2-2x+y=4x^2+4y^2+4y+1$
$\Rightarrow 2x^2+2y^2+2x+3y+1=0$
$\Rightarrow x^2+y^2+x+\frac{3}{2}y+\frac{1}{2}=0$

Circle with center $\left(-\frac{1}{2}, -\frac{3}{4}\right)$
$r=\sqrt{\frac{1}{4}+\frac{9}{16}-\frac{1}{2}}=\sqrt{\frac{4+9-8}{16}}=\frac{\sqrt{5}}{4} \Rightarrow d=\frac{\sqrt{5}}{2}$

where, $d=$diameter of the given circle.