Explanation

(b)
Given equation of circles are
$x^{2} + y^{2} + 4x + 8y = 0$ …(i)
and $x^{2} + y^{2} + 8x + 2ky = 0$ …(ii)
for circle (i),
centre, $C_{1} = ( - 2,\ - 4)$
and radius $r_{1} = \sqrt{4 + 16 - 0} = \sqrt{20} = 2\sqrt{5}$
for circle (ii),
centre, $C_{2} = ( - 4,\ - k)$
and radius, $r_{2} = \sqrt{16 + k^{2} - 0} = \sqrt{16 + k^{2}}$
given circles touch each other externally
$\therefore\ \ \ \left| C_{1}C_{2} \right| = r_{1} + r_{2}$
$\Rightarrow \sqrt{( - 2 + 4)^{2} + ( - 4 + k)^{2}} = 2\sqrt{5} + \sqrt{16 + k^{2}}$
$\Rightarrow \ \ 4 + 16 + k^{2} - 8k = 20 + 16 + k^{2} + 4\sqrt{5}\sqrt{16 + k^{2}}$
$\Rightarrow \ \ - 16 - 8k = 4\sqrt{5}\sqrt{16 + k^{2}}$
$\Rightarrow \ - 4 - 2k = \sqrt{5}\sqrt{16 + k^{2}}$
$\Rightarrow (4 + 2k) = \left( \sqrt{5}\sqrt{16 + k^{2}} \right)^{2}$
$\Rightarrow \ \ 16 + 4k^{2} + 16k = 5\left( 16 + k^{2} \right)$
$\Rightarrow \ \ k^{2} - 16k + 64 = 0$
$\Rightarrow (k - 8)^{2} = 0$
$\Rightarrow \ \ k = 8$