JEE Main 2024 — 9 April, Shift 1 — Question 22

$|z-1| \leq 1$ $\Rightarrow|(\mathrm{x}-1)+\mathrm{iy}| \leq 1$

$\Rightarrow \sqrt{(\mathrm{x}-1)^{2}+\mathrm{y}^{2}} \leq 1$

$\Rightarrow(\mathrm{x}-1)^{2}+\mathrm{y}^{2} \leq 1$

$\qquad$ Also $|z-5| \leq|z-5 i|$

$(x-5)^{2}+y^{2} \leq x^{2}+(y-5)^{2}$

$-10 x \leq-10 y$ $\Rightarrow x \geq y$ Solving (1) and (2)

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

$\Rightarrow 2 \mathrm{x}^{2}-2 \mathrm{x}=0$

$\Rightarrow \mathrm{x}(\mathrm{x}-1)=0$

$\Rightarrow \mathrm{x}=0$ or $\mathrm{x}=1$

$\mathrm{y}=0 \text { or } \mathrm{y}=1$

Given $\mathrm{x}, \mathrm{y} \in \mathrm{I}$

Points $(0,0),(1,0),(2,0),(1,1),(1,-1)$ to find $\left|\mathrm{z}_{1}\right|^{2}+\left|\mathrm{z}_{2}\right|^{2}+\left|\mathrm{z}_{3}\right|^{2}+\left|\mathrm{z}_{4}\right|^{2}+\left|\mathrm{z}_{5}\right|^{2}$

$=0+1+4+2+2=9$