Answer : $k=\pm 8$
Equation of the line which touches the circle is $y=\sqrt 3x+k\Rightarrow \sqrt 3x-y+k=0$
Equation of the circle is $x^2+y^2=16$
$\Rightarrow$ The coordinate of the origin is $(0,0)$, radius $'r'=4$.
The length of the perpendicular from $(0,0)$ to the line is the radius.
$\therefore 4=\bigg|\large\frac{\sqrt 3(0)-(0)+k}{\sqrt{(\sqrt 3)^2+(1)^2}}\bigg|$
$\Rightarrow 4 = \large\frac{k}{2}$
$\therefore k =\pm 8$