Complex Numbers

Hardmathematics

If |z - 2| = 2|z - 1|, then the locus of z is

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A circle

A straight line

A parabola

An ellipse

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About This Question

Subject: mathematics
Chapter: complex numbers and quadratic equations
Topic: complex numbers
Difficulty: Hard
Year: 2025

Solution

Correct Answer:

A circle

Assumption: Let $z = x + i y$ . Then $∣ z - 2∣ = (x - 2)^{2} + y^{2}$ and $∣ z - 1∣ = (x - 1)^{2} + y^{2}$ .
Equation: $∣ z - 2∣ = 2∣ z - 1∣ ⟹ ∣ z - 2 ∣^{2} = 4∣ z - 1 ∣^{2}$ .
Expansion:
- $(x - 2)^{2} + y^{2} = 4 [(x - 1)^{2} + y^{2}]$
- $x^{2} - 4 x + 4 + y^{2} = 4 (x^{2} - 2 x + 1 + y^{2})$
- $x^{2} - 4 x + 4 + y^{2} = 4 x^{2} - 8 x + 4 + 4 y^{2}$
Rearrange:
- $3 x^{2} + 3 y^{2} - 4 x = 0$
- $x^{2} + y^{2} - \frac{4}{3} x = 0$
Completion: $(x - \frac{2}{3})^{2} + y^{2} = (\frac{2}{3})^{2}$ . This is the standard equation of a circle with center $(2/3, 0)$ and radius $2/3$ .

This hard difficulty mathematics question is from the chapter complex numbers and quadratic equations, covering the topic of complex numbers. It appeared in the 2025 exam.

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