[UPLIFTJEE]

Let $\mathrm{S}=\left\{\mathrm{z} \in \mathrm{C}:\left|\frac{\mathrm{z}-6 \mathrm{i}}{\mathrm{z}-2 \mathrm{i}}\right|=1\right.$ and $\left.\left|\frac{\mathrm{z}-8+2 \mathrm{i}}{\mathrm{z}+2 \mathrm{i}}\right|=\frac{3}{5}\right\}$. Then $\sum_{\mathrm{z} \in \mathrm{s}}|\mathrm{z}|^{2}$ is equal to [JEE MAIN 24 Jan 2026 S I]
(A) 398
(B) 413
(C) 423
(D) 385

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[UPLIFTJEE]

Solution

$\left|\frac{\mathrm{z}-6 \mathrm{i}}{\mathrm{z}-2 \mathrm{i}}\right|=1$ let $\mathrm{z}=\mathrm{x}+\mathrm{iy} \Rightarrow|\mathrm{z}-6 \mathrm{i}|=|\mathrm{z}-2 \mathrm{i}|$

\[ \begin{aligned} & \Rightarrow|x+i y-6 i|=|x+i y-2 i| \Rightarrow x^{2}+(y-6)^{2}=x^{2}+(y-2)^{2} \\ & \Rightarrow y^{2}+36-12 y=y^{2}+4-4 y \Rightarrow 8 y=32 \Rightarrow y=4 \\ & z=x+4 i ; \frac{|z-8+2 i|}{|z+2 i|}=\frac{3}{5} \Rightarrow|x+4 i-8+2 i|=\frac{3}{5}|x+4 i+2 i| \\ & \Rightarrow|x-8+6 i|=\frac{3}{5}|x+6 i| \Rightarrow \sqrt{(x-8)^{2}+36}=\frac{3}{5} \sqrt{x^{2}+36} \Rightarrow(x-8)^{2}+36=\frac{9}{25}\left(x^{2}+36\right) \\ & \Rightarrow 25\left(x^{2}-16 x+100\right)=9 x^{2}+324 \Rightarrow(x-17)(x-8)=0 \Rightarrow x=17,8 \\ & z=17+4 i, z=8+4 i \Rightarrow \sum_{z \in S}|z|^{2}=(289+16)+(64+16)=305+80=385 \end{aligned} \]

[UPLIFTJEE]

Answer

D

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