Explanation

(b)
Equation of the coaxial system of circle is
$S_{1} + \lambda S_{2} = 0$
$\therefore\ \ \left( x^{2} + y^{2} + 5x + y + 4 \right) + \lambda\left( x^{2} + y^{2} + 10x - 4y - 1 \right) = 0$
$\Rightarrow \ \ x^{2} + y^{2} + \frac{5(1 + 2\lambda)}{(1 + \lambda)}x + \frac{(1 - 4\lambda)}{(1 + \lambda)}y + \frac{4 - \lambda}{1 + \lambda} = 0$
$\therefore$ The centre of the circle is
$\left( - \frac{5(1 + 2\lambda)}{2(1 + \lambda)},\ - \frac{(1 - 4\lambda)}{2(1 + \lambda)} \right)$…(i)
For limiting point, $r = 0$
$\therefore\ \ \sqrt{\frac{25(1 + 2\lambda)^{2}}{4(1 + \lambda)^{2}} + \frac{(1 - 4\lambda)^{2}}{4(1 + \lambda)^{2}} - \frac{(4 - \lambda)}{(1 + \lambda)}} = 0$
$\Rightarrow \ \ 25(1 + 2\lambda)^{2} + (1 - 4\lambda)^{2} - 4(4 - \lambda)(1 + \lambda) = 0$
$\Rightarrow \ \ 120\lambda^{2} + 80\lambda + 10 = 0\ \Rightarrow \ (6\lambda + 1)(2\lambda + 1) = 0$
$\Rightarrow \ \lambda = - \frac{1}{6}$ and $- \frac{1}{2}$
On substituting the values of $\lambda$ in Eq. (i), we get $( - 2,\ - 1)$ and $(0,\ - 3)$