Explanation

(b)
Let the equation of circles be
$S_{1} \equiv x^{2} + y^{2} + 13x - 3y = 0$ …(i)
And $S_{2} = 2x^{2} + 2y^{2} + 4x - 7y - 25 = 0$ …(ii)
The equation of intersecting circle is $\lambda S_{1} + S_{2} = 0$
$\Rightarrow \ \lambda\left( x^{2} + y^{2} + 13x - 3y \right) + \left( x^{2} + y^{2} + 2x - \frac{7y}{2} - \frac{25}{2} \right) = 0$
$\Rightarrow \ \left\lbrack x^{2}(1 + \lambda) + y^{2}(1 + \lambda) + x(2 + 13\lambda) - y\left( \frac{7}{2} + 3\lambda \right) - \frac{25}{2} \right\rbrack = 0$ …(iii)
$\therefore\ Centre = \left( - \frac{(2 + 13\lambda)}{2(1 + \lambda)},\frac{(7/2) + 3\lambda}{2(1 + \lambda)} \right)$
$\because Centre\ lies\ on\ 13x + 30y = 0\$
$\Rightarrow \ - 13\left( \frac{2 + 13\lambda}{2} \right) + 30\left( \frac{(7/2) + 3\lambda}{2} \right) = 0$
$\Rightarrow \ - 26 - 169\lambda + 105 + 90\lambda = 0\ \Rightarrow \ \lambda = 1$
Hence, putting the value of $x$ in Eq. (iii), then required equation of circle is
$4x^{2} + 4y^{2} + 30x - 13y - 25 = 0$