JEE Main 2025 — 7 April, Evening Shift — Question 30

$\operatorname{Re}(z)+\operatorname{Re}(\bar{z})=2$

$\Rightarrow \operatorname{Re}(z)=1$

$\Rightarrow \operatorname{Re}\left(\frac{x+i y-1}{2 x+i(2 y+1)}\right)=1$

$\Rightarrow \operatorname{Re}\left(\frac{(x-1)+i y}{2 x+i(2 y+1)}\right)=1$

$\Rightarrow \operatorname{Re} \frac{((x-1)+i y)(2 x-i(2 y+1))}{(2 x+i(2 y+1))(2 x-i(2 y+1))}=1$

$\Rightarrow \frac{2 x^{2}-2 x+2 y^{2}+y}{4 x^{2}+(2 y+1)^{2}}=1$

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

Centre $(a, b)=\left(-\frac{1}{2}, \frac{-3}{4}\right)$

Radius $=\frac{\sqrt{5}}{4}$

$\frac{15 a b}{r^{2}}=15 \cdot \frac{3}{8} \times \frac{16}{5}=18$