Question 7

Let $z = x + iy$. The condition $|z - z_1| = 2|z - z_2|$ implies $|z - (1 + 2i)|^2 = 4|z - 3i|^2$.

This expands to $(x-1)^2 + (y-2)^2 = 4[x^2 + (y-3)^2]$.

Expanding and simplifying: $x^2 - 2x + 1 + y^2 - 4y + 4 = 4x^2 + 4y^2 - 24y + 36 \Rightarrow 3x^2 + 3y^2 + 2x - 20y + 31 = 0$.

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

The centre is $(-\frac{1}{3}, \frac{10}{3})$ and the radius $r = \sqrt{(\frac{1}{3})^2 + (\frac{10}{3})^2 - \frac{31}{3}} = \sqrt{\frac{1 + 100 - 93}{9}} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}$.

Thus (A) and (D) are correct.