Locus of Complex Points Representing Apollonius Circle

To determine the locus of the points in the set $S$, let $z = x + iy$, where $x, y \in \mathbb{R}$.

Given the complex numbers: $z_1 = 1 + 2i \quad \text{and} \quad z_2 = 3i$

The given condition is: $|z - z_1| = 2|z - z_2|$

Substituting $z = x + iy$, $z_1 = 1 + 2i$, and $z_2 = 3i$: $|(x - 1) + i(y - 2)| = 2|x + i(y - 3)|$

Squaring both sides to eliminate the modulus: $(x - 1)^2 + (y - 2)^2 = 4\left[x^2 + (y - 3)^2\right]$

Expanding both sides: $x^2 - 2x + 1 + y^2 - 4y + 4 = 4\left(x^2 + y^2 - 6y + 9\right)$ $x^2 + y^2 - 2x - 4y + 5 = 4x^2 + 4y^2 - 24y + 36$

Rearranging all terms to one side: $3x^2 + 3y^2 + 2x - 20y + 31 = 0$

Dividing by $3$: $x^2 + y^2 + \frac{2}{3}x - \frac{20}{3}y + \frac{31}{3} = 0$

This is the standard equation of a circle of the form $x^2 + y^2 + 2gx + 2fy + c = 0$, where: $g = \frac{1}{3}, \quad f = -\frac{10}{3}, \quad c = \frac{31}{3}$

1. Centre of the circle: $\text{Centre} = (-g, -f) = \left(-\frac{1}{3}, \frac{10}{3}\right)$

2. Radius of the circle: $r = \sqrt{g^2 + f^2 - c} = \sqrt{\left(\frac{1}{3}\right)^2 + \left(-\frac{10}{3}\right)^2 - \frac{31}{3}}$ $r = \sqrt{\frac{1}{9} + \frac{100}{9} - \frac{93}{9}} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}$

Thus:

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

Therefore, the correct options are A and D.