Q37 / 75 · JEE Main 2026 (24 January Shift 1) · Mathematics
Let $\mathrm{S}=\left\{z \in \mathbb{C}:\left|\frac{z-6 i}{z-2 i}\right|=1\right.$ and $\left.\left|\frac{z-8+2 i}{z+2 i}\right|=\frac{3}{5}\right\}$.
Then $\sum_{z \in \mathrm{~s}}|z|^{2}$ is equal to
Then $\sum_{z \in \mathrm{~s}}|z|^{2}$ is equal to
- Correct answerA385
- B398
- C413
- D423
Solution
Solving $\left| \dfrac{z - 6i}{z - 2i} \right| = 1 \Rightarrow y = 4$ .....(1)
(where $z = x + iy$)
Now solving $\left| \dfrac{z - 8 + 2i}{z + 2i} \right| = \dfrac{3}{5}$
$\Rightarrow x^2 + y^2 - 25x + 4y + 104 = 0$ .....(2)
Solving (1) & (2) $\Rightarrow z = 17 + 4i$ & $8 + 4i$
$\Rightarrow \sum |z|^2 = (17)^2 + (4)^2 + (8)^2 + (4)^2 = 385$
(where $z = x + iy$)
Now solving $\left| \dfrac{z - 8 + 2i}{z + 2i} \right| = \dfrac{3}{5}$
$\Rightarrow x^2 + y^2 - 25x + 4y + 104 = 0$ .....(2)
Solving (1) & (2) $\Rightarrow z = 17 + 4i$ & $8 + 4i$
$\Rightarrow \sum |z|^2 = (17)^2 + (4)^2 + (8)^2 + (4)^2 = 385$
Hence, the correct answer is option: 'A'.