If the line y = √3x + k touches the circle

Since, the equation of a circle having centre $(h, k)$, having radius as $r$ units, is

$(x-h)^{2}+(y-k)^{2}=r^{2}$

$(x-0)^{2}+(y-0)^{2}=4^{2}$

Perpendicular Distance between a point $(0,0) \&$ the line

$y=\sqrt{3} x+k$ or $\sqrt{3} x-y+k=0$

Perpendicular Distance (Between a point and line) $=$

Whereas the point is $\left(x_{1}, y_{1}\right)$ and the line is expressed as $a x+b y+c=0$

$D=\frac{\{|\sqrt{3}(0)+(0)(-1)+k|\}}{\sqrt{(\sqrt{3})^{2}+1^{2}}}=\frac{\{|k|\}}{\sqrt{3+1}}=\frac{\{k\}}{\sqrt{4}}=\frac{k}{2} \quad \frac{\left|a x_{1}+b y_{1}+c\right|}{\sqrt{a^{2}+b^{2}}}$

$\frac{k}{2}=4$ (Radius $=4$, Given)

$k=8$

Hence, the required value of $k$ is $8 .$

Hence, the required value of k is 8.