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?

$z_1 = 1 + 2i,\quad z_2 = 3i$ $S = \{(x, y) : |x + iy - z_1| = 2|x + iy - z_2|\}$ $\text{To find: the centre and radius of the circle } S.$ $\text{Plan:}$ $1.\ \text{moduli} \to \text{distances};\quad 2.\ \text{square};\quad 3.\ \text{centre and radius}$ $A = z_1 = (1, 2),\quad B = z_2 = (0, 3)$ $|x + iy - z_1| = PA,\quad |x + iy - z_2| = PB$ $PA = 2\,PB$ $(x-1)^2 + (y-2)^2 = 4\left[\,x^2 + (y-3)^2\,\right]$ $x^2 - 2x + 1 + y^2 - 4y + 4 = 4x^2 + 4y^2 - 24y + 36$ $3x^2 + 3y^2 + 2x - 20y + 31 = 0$ $x^2 + y^2 + \dfrac{2}{3}x - \dfrac{20}{3}y + \dfrac{31}{3} = 0$ $2g = \dfrac{2}{3} \Rightarrow g = \dfrac{1}{3},\quad 2f = -\dfrac{20}{3} \Rightarrow f = -\dfrac{10}{3}$ $c = \dfrac{31}{3}$ $\text{Centre} = (-g, -f) = \left(-\dfrac{1}{3}, \dfrac{10}{3}\right)$ $r = \sqrt{g^2 + f^2 - c} = \sqrt{\dfrac{1}{9} + \dfrac{100}{9} - \dfrac{31}{3}}$ $r = \sqrt{\dfrac{8}{9}} = \dfrac{2\sqrt{2}}{3}$ $\text{Centre} \left(-\dfrac{1}{3},\dfrac{10}{3}\right):\ (1)\ \checkmark,\ (2)\ \times$ $\text{Radius} \dfrac{2\sqrt{2}}{3}:\ (4)\ \checkmark,\ (3)\ \times$ $\boxed{(1)\ \text{and}\ (4)\ \text{are TRUE}}$