Let $\mathbb{R}$ denote the set of all real numbers. Let $z_{1}=1+2 i$ and $z_{2}=3 i$ be two complex numbers, where $i=\sqrt{-1}$. Let \[ S=\left\{(x, y) \in \mathbb{R} \times \mathbb{R}:\left|x+i y-z_{1}\right|=2\left|x+i y-z_{2}\right|\right\} . \] Then which of the following statements is (are) TRUE ?

(A) $S$ is a circle with centre $\left(-\frac{1}{3}, \frac{10}{3}\right)$; (D) $S$ is a circle with radius $\frac{2 \sqrt{2}}{3}$

(B) $S$ is a circle with centre $\left(\frac{1}{3}, \frac{8}{3}\right)$

(C) $S$ is a circle with radius $\frac{\sqrt{2}}{3}$

JEE Advanced · 2025 · 18 May 2025 · Paper 1 · Complex Numbers

Let

R

denote the set of all real numbers. Let

z_{1} = 1 + 2 i

and

z_{2} = 3 i

be two complex numbers, where

i = - 1

. Let

S = {(x, y) \in R \times R : ∣ x + i y - z_{1} ∣ = 2 ∣ x + i y - z_{2} ∣} .

Then which of the following statements is (are) TRUE ?

Answer: (A), (D)

The condition

∣ x + i y - 1 - 2 i ∣ = 2∣ x + i y - 3 i ∣

becomes:

(x - 1)^{2} + (y - 2)^{2} = 4 [x^{2} + (y - 3)^{2}]

Expanding:

x^{2} - 2 x + 1 + y^{2} - 4 y + 4 = 4 x^{2} + 4 y^{2} - 24 y + 36

3 x^{2} + 3 y^{2} + 2 x - 20 y + 31 = 0

x^{2} + y^{2} + \frac{2 x}{3} - \frac{20 y}{3} + \frac{31}{3} = 0

Centre:

(- \frac{1}{3}, \frac{10}{3})

. (A) is true, (B) is false.

Radius:

r = \frac{1}{9} + \frac{100}{9} - \frac{31}{3} = \frac{8}{9} = \frac{2 2}{3}

. (D) is true, (C) is false.

JEE Advanced · 2025 · 18 May 2025 · Paper 1 · One or more correct options