complex numbers

Question

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

RankGuru · Accepted Answer

1. **Assumption**: Let $z = x + iy$. Then $|z-2| = \sqrt{(x-2)^2 + y^2}$ and $|z-1| = \sqrt{(x-1)^2 + y^2}$.
2. **Equation**: $|z-2| = 2|z-1| \implies |z-2|^2 = 4|z-1|^2$.
3. **Expansion**:
   - $(x-2)^2 + y^2 = 4[(x-1)^2 + y^2]$
   - $x^2 - 4x + 4 + y^2 = 4(x^2 - 2x + 1 + y^2)$
   - $x^2 - 4x + 4 + y^2 = 4x^2 - 8x + 4 + 4y^2$
4. **Rearrange**:
   - $3x^2 + 3y^2 - 4x = 0$
   - $x^2 + y^2 - \frac{4}{3}x = 0$
5. **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 $2/3$.

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