Explanation

(b)
Since, $\left| \frac{z - i}{z + i} \right| = 2\ \ \ \ \ \Rightarrow \ \ \ \ \left| \frac{x + iy - i}{x + iy + i} \right| = 2$ [where $z = x + iy$]
$\Rightarrow \ \ \ \ \ \left| x + i(y - 1) \right| = 2|x + (y + 1)i|$
$\Rightarrow \ \ \ \ \ x^{2} + (y - 1)^{2} = 4\lbrack x^{2} + (y + 1)^{2}\rbrack$
$\Rightarrow \ \ \ \ \ x^{2} + y^{2} - 2y + 1 = 4x^{2} + 4y^{2} + 8y + 4$
$\Rightarrow \ \ \ 3x^{2} + 3y^{2} + 10y + 3 = 0$