Explanation

(d)
Let the general equation of circle be
$x^{2} + y^{2} + 2gx + 2fy + c = 0$ …(i)
It cuts the circle $x^{2} + y^{2} - 20x + 4 = 0$ orthogonally, then
By the condition, $2\left( g_{1}g_{2} + f_{1}f_{2} \right) = c_{1} + c_{2}$
$2( - 10g + 0 \times f) = c + 4\ \Rightarrow \ - 20g = c + 4$ …(ii)
$\because$ Circle (i) touches the line $x = 2$ or $x + 0y - 2 = 0$
$\therefore$ Perpendicular distance from centre to the tangent=radius
$\Rightarrow \ \left| \frac{- g + 0 - 2}{\sqrt{1^{2} + 0^{2}}} \right| = \sqrt{g^{2} + f^{2} - c}$
$\Rightarrow (g + 2)^{2} = g^{2} + f^{2} - c$
$\Rightarrow g^{2} + 4 + 4g = g^{2} + f^{2} - c$
$\Rightarrow 4g + 4 = f^{2} - c$ …(iii)
On eliminating $c$ from Eqs. (ii) and (iii), we get
$- 16g + 4 = f^{2} + 4\ \Rightarrow f^{2} + 16g = 0$
Hence, the locus of $( - g,\ - f)$ is $y^{2} - 16x = 0$
(replacing $–f$ and $–g$ by $x$ and $y$)