The equation of a common tangent to the circle ${\mathrm{x}}^2+{\mathrm{y}}^2=16$ and to the ellipse $\frac{{\mathrm{x}}^2}{49}+\frac{{\mathrm{y}}^2}{4}=1$ is

Given ellipse $\frac{{\mathrm{x}}^2}{49}+\frac{{\mathrm{y}}^2}{4}=1$

Let $\mathrm{y}=\mathrm{mx}+\mathrm{c}$ be tangent to ellipse then 

$\begin{array}{rcl}{\mathrm{c}}^2& =& {\mathrm{a}}^2{\mathrm{m}}^2+{\mathrm{b}}^2\\ & \Rightarrow & \mathrm{y}=mx\pm \sqrt{{\mathrm{a}}^2{\mathrm{m}}^2+{\mathrm{b}}^2}\\ & \Rightarrow & \mathrm{y}=\mathrm{mx}\pm \sqrt{49{\mathrm{m}}^2+4}\end{array}$

It is also tangent to the circle ${\mathrm{x}}^2+{\mathrm{y}}^2=16$ then 

$\begin{array}{rcl} & \Rightarrow & \mathrm{r}=\mathrm{d} \mathrm{where} \mathrm{r}=4, C=(0, 0)\\ & \Rightarrow & 4=\frac{\left|\sqrt{49{\mathrm{m}}^2+4}\right|}{{\mathrm{m}}^2+1}\\ & \Rightarrow & 16\left({\mathrm{m}}^2+1\right)=49{\mathrm{m}}^2+4\\ & \Rightarrow & 12=33{\mathrm{m}}^2\\ & \Rightarrow & {\mathrm{m}}^2=\frac{12}{33}=\frac{4}{11}\\ & \Rightarrow & \mathrm{m}=\frac{2}{\sqrt{11}}\end{array}$

$\therefore$ Common tangent is 

$\begin{array}{rcl}\mathrm{y}& =& \frac{2}{\sqrt{11}}\mathrm{x}\pm \sqrt{49\left(\frac{4}{11}\right)+4}\\ & \Rightarrow & \mathrm{y}=\frac{2}{\sqrt{11}}\mathrm{x}\pm \frac{4\sqrt{15}}{\sqrt{11}}\\ & \Rightarrow & \sqrt{11} \mathrm{y}=2\mathrm{x}\pm 4\sqrt{15}\end{array}$