JEE Main 2025 — 4 April, Morning Shift — Question 42

Let $z=x+i y$ $|z-2-i|=3 \Rightarrow(x-2)^{2}+(y-1)^{2}=3^{2}$

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

$\Rightarrow A=\left\{(x, y):(x-2)^{2}+(y-1)^{2}=3^{2}, x, y \in R\right\}$,

$B=\{(x, y): x+y=2\}$

$\Rightarrow x-2=-y \Rightarrow y^{2}+(y-1)^{2}=3^{2}$

$\Rightarrow 2 y^{2}-2 y-8=0 \Rightarrow y^{2}-y-4=0$

$y_{1}+y_{2}=1, y_{1} y_{2}=-4$

$\Rightarrow y_{1}^{2}+y_{2}^{2}$

$=\left(y_{1}+y_{2}\right)^{2}-2 y_{1} y_{2}=9$

$\Rightarrow x_{1}+x_{2}=4\left(y_{1}+y_{2}\right)=3$,

$x_{1} x_{2}=\left(2-y_{1}\right)\left(2-y_{2}\right)=4-2\left(y_{1}+y_{2}\right)+y_{1} y_{2}=-2$

$\Rightarrow x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=13$

$\because S=\left\{\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)\right\}$

$\Rightarrow \sum_{z \in S}|z|^{2}=\left(x_{1}^{2}+y_{1}^{2}\right)+\left(x_{2}^{2}+y_{2}^{2}\right)=22$