Explanation

(a)
Given $\arg{\left( \frac{z - 2}{z - 6i} \right) = \frac{\pi}{2}}$
$\therefore\ \ \arg{(z - 2) - \arg{(z - 6i}) = \frac{\pi}{2}}$
$\Rightarrow \ \arg{\left\lbrack (x - 2) + iy \right\rbrack - \arg{\left\lbrack x + i(y - 6) \right\rbrack = \frac{\pi}{2}}}$
$\Rightarrow \ \ \tan^{- 1}{\frac{y}{x - 2} - \tan^{- 1}{\frac{y - 6}{x} = \frac{\pi}{2}}}$
$\Rightarrow \ \ \ \ \left( \frac{\frac{y}{x - 2} - \frac{y - 6}{x}}{1 + \frac{y}{x - 2}.\frac{y - 6}{x}} \right) = \tan\frac{\pi}{2}$
$\Rightarrow \ \ \ \ 1 + \frac{y}{x - 2}.\frac{y - 6}{x} = 0$
$\Rightarrow \ \ \ x(x - 2) + y(y - 6) = 0$
This is an equation of circle in diametric form.