Explanation

(b)
Let the equation of circle passing through origin is
$x^{2} + y^{2} + 2gx + 2fy = 0$
It also passes through (2, 1)
$\therefore\ \ \ 4 + 1 + 4g + 2f = 0$
$\Rightarrow \ 4g + 2f = - 5$ …(i)
Also, circle touches the line $y = x$
$\therefore\ \$Perpendicular from centre$( - g,\ - f)$ to the tangent=radius
$\Rightarrow \ \ \frac{| - f + g|}{\sqrt{1^{2} + 1^{2}}} = \sqrt{g^{2} + \ f^{2}} \Rightarrow \ \ f^{2} + g^{2} - 2fg = 2\left( g^{2} + f^{2} \right)$
$\Rightarrow \ \ \ (g + f)^{2} = 0\ \ \Rightarrow g = - f$
$\therefore$ From Eq. (i), $4( - f) + 2f = - 5$
$\Rightarrow \ \ \ f = \frac{5}{2}$ and $g = - \frac{5}{2}$
$\therefore\ \ \ x^{2} + y^{2} - 5x + 5y = 0$
On comparing with $x^{2} + \ y^{2} + px + qy = 0$
$\therefore\ \ p = - 5,\ q = 5$