Explanation

(b)
Since, the given circles cut each other orthogonally
$\therefore g_{1}g_{2} + a^{2} = 0$ …(i)
If $lx + my = 1$ is a common tangent of these circles, then
$\frac{- lg_{1} - 1}{\sqrt{l^{2} + m^{2}}} = \pm \sqrt{g_{1}^{2} + a^{2}}$
$\Rightarrow \left( lg_{1} + 1 \right)^{2} = \left( l^{2} + m^{2} \right)\left( g_{1}^{2} + a^{2} \right)$
$\Rightarrow m^{2}g_{1}^{2} - 2lg_{1} + a^{2}\left( l^{2} + m^{2} \right) - 1 = 0$
Similarly, $m^{2}g_{2}^{2} - 2lg_{2} + a^{2}\left( l^{2} + m^{2} \right) - 1 = 0$
So, that $g_{1},g_{2}$ are the roots of the equation
$m^{2}g^{2} - 2\lg + a^{2}\left( l^{2} + m^{2} \right) - 1 = 0$
$\Rightarrow \ \ g_{1}g_{2} = \frac{a^{2}\left( l^{2} + m^{2} \right) - 1}{m^{2}} = - a^{2}$ [from Eq. (i)]
$\Rightarrow a^{2}\left( l^{2} + m^{2} \right) = 1 - a^{2}m^{2}$ …(ii)
Now, $p_{1}p_{2} = \frac{|ma - 1|}{\sqrt{l^{2} + m^{2}}}.\frac{| - ma - 1|}{\sqrt{l^{2} + m^{2}}}$
$= \frac{|1 - m^{2}a^{2}|}{l^{2} + m^{2}} = a^{2}$ [from Eq. (ii)]