Complex Numbers
PYP_JEE_ADV_2025_P1
Grade None
Question:
Let $\mathbb{R}$ denote the set of all real numbers. Let $z_1 = 1 + 2i$ and $z_2 = 3i$ be two complex numbers, where $i = \sqrt{-1}$. Let
$$S = \{(x, y) \in \mathbb{R} \times \mathbb{R} : |x + iy - z_1| = 2|x + iy - z_2|\}.$$
Then which of the following statements is (are) TRUE?
$S$ is a circle with centre $\left(-\dfrac{1}{3}, \dfrac{10}{3}\right)$
$S$ is a circle with centre $\left(\dfrac{1}{3}, \dfrac{8}{3}\right)$
$S$ is a circle with radius $\dfrac{\sqrt{2}}{3}$
$S$ is a circle with radius $\dfrac{2\sqrt{2}}{3}$
Step-by-Step Solution
Key Concept: Apollonius circle: locus of points where ratio of distances to two fixed points is constant
Let $z = x+iy$. The condition $|z - z_1| = 2|z - z_2|$ gives:
$(x-1)^2 + (y-2)^2 = 4[(x-0)^2 + (y-3)^2]$
$x^2 - 2x + 1 + y^2 - 4y + 4 = 4x^2 + 4y^2 - 24y + 36$
$3x^2 + 3y^2 + 2x - 20y + 31 = 0$
$x^2 + y^2 + \dfrac{2}{3}x - \dfrac{20}{3}y + \dfrac{31}{3} = 0$
Centre: $\left(-\dfrac{1}{3}, \dfrac{10}{3}\right)$. ✓ (A is TRUE, B is FALSE)
Radius$^2 = \dfrac{1}{9} + \dfrac{100}{9} - \dfrac{31}{3} = \dfrac{101}{9} - \dfrac{93}{9} = \dfrac{8}{9}$.
Radius $= \dfrac{2\sqrt{2}}{3}$. ✓ (D is TRUE, C is FALSE)
Correct Answer: A, D