Explanation

(c)
The centres and radii of given circles are $C_{1}( - 1,\ - 4),\ C_{2}(2,\ 5)$
and $r_{1} = \sqrt{1 + 16 + 23} = \sqrt{40}$,
$r_{2} = \sqrt{4 + 25 - 9} = \sqrt{20}$
Now, $C_{1}C_{2} = \sqrt{(2 + 1)^{2} + (5 + 4)^{2}} = \sqrt{90}$
And $r_{1} + r_{2} = \sqrt{40} + \sqrt{20}$
Here, $C_{1}C_{2} < r_{1} + r_{2}$
$\therefore$ Two common tangents can be drawn