JEE Advanced 2025 Paper 1 Q07 Mathematics Algebra Complex Numbers Medium

JEE Advanced 2025 Paper 1 · Q07 · Complex Numbers

Let $z_1 = 1 + 2i$ and $z_2 = 3i$ (where $i = \sqrt{-1}$). Let $S = \{ (x, y) \in \mathbb{R} \times \mathbb{R} : |x + iy - z_1| = 2|x + iy - z_2| \}$. Which statements about $S$ are TRUE?

  1. A. $S$ is a circle with centre $\left(-\frac{1}{3}, \frac{10}{3}\right)$
  2. B. $S$ is a circle with centre $\left(\frac{1}{3}, \frac{8}{3}\right)$
  3. C. $S$ is a circle with radius $\frac{\sqrt{2}}{3}$
  4. D. $S$ is a circle with radius $\frac{2\sqrt{2}}{3}$
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:A, D

Solution

Let $z = x + iy$. $|z - z_1|^2 = 4|z - z_2|^2$: $(x-1)^2 + (y-2)^2 = 4[x^2 + (y-3)^2]$. Expand: $x^2 - 2x + 1 + y^2 - 4y + 4 = 4x^2 + 4y^2 - 24y + 36$. $0 = 3x^2 + 3y^2 + 2x - 20y + 31$. Divide by $3$: $x^2 + y^2 + \frac{2}{3}x - \frac{20}{3}y + \frac{31}{3} = 0$. Centre $= \left(-\frac{1}{3}, \frac{10}{3}\right)$. (A) TRUE; (B) FALSE. Radius$^2 = \frac{1}{9} + \frac{100}{9} - \frac{31}{3} = \frac{101 - 93}{9} = \frac{8}{9}$. $r = \frac{2\sqrt{2}}{3}$. (D) TRUE; (C) FALSE. Answer: A, D.

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