If the locus of z ∈ C, such that Re ((z-1)/(2 z+i))+ Re ((z̄-1)/(2 z̄-i))=2, is a circle of radius r and center ( a, b ), then (15 ab)/(r^2) is equal…

If the locus of $\displaystyle z \in \mathrm{C}$, such that $\displaystyle \operatorname{Re}\left(\frac{z-1}{2 z+i}\right)+\operatorname{Re}\left(\frac{\bar{z}-1}{2 \bar{z}-i}\right)=2$, is a circle of radius r and center $\displaystyle (\mathrm{a}, \mathrm{b})$, then $\displaystyle \frac{15 \mathrm{ab}}{\mathrm{r}^2}$ is equal to :