JEE Main 2026 — Jan 24, Shift 1 — Mathematics solution

JEE Main2026Jan 24, Shift 1Mathematics

Let $S = {z \in C : \frac{z - 6 i}{z - 2 i} = 1 and \frac{z - 8 + 2 i}{z + 2 i} = \frac{3}{5}}$ . Then $z \in S \sum ∣ z ∣^{2}$ is equal to

Solution

Let $z = x + i y$ .

First equation: $∣ z - 6 i ∣ = ∣ z - 2 i ∣$ — perpendicular bisector of $6 i$ and $2 i$ , giving $y = 4$ .

Second equation: $5∣ z - 8 + 2 i ∣ = 3∣ z + 2 i ∣$ . Squaring with $y = 4$ (so $y + 2 = 6$ ):

$25 [(x - 8)^{2} + 36] = 9 [x^{2} + 36]$

$25 x^{2} - 400 x + 1600 + 900 = 9 x^{2} + 324$

$16 x^{2} - 400 x + 2176 = 0 \Rightarrow x^{2} - 25 x + 136 = 0$ .

Roots: $x = \frac{25 \pm 625 - 544}{2} = \frac{25 \pm 9}{2} = 17 or 8$ .

So $z = 17 + 4 i$ or $z = 8 + 4 i$ .

$\sum ∣ z ∣^{2} = (1 7^{2} + 16) + (8^{2} + 16) = 305 + 80 = 385$ .

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