Explanation

(d)
Let the equation of tangent to the circle $x^{2} + y^{2} = 16$ is
$y = mx + 4\sqrt{1 + m^{2}}$ $\left( \because y = mx + a\sqrt{1 + m^{2}} \right)$
And let the equation of tangent to the ellipse $\frac{x^{2}}{49} + \frac{y^{2}}{4} = 1$ is
$y = mx + \sqrt{49m^{2} + 4}$ $\left( \because y = mx + \sqrt{a^{2}m^{2} + b^{2}} \right)$
For common tangent
$4\sqrt{1 + m^{2}} = \sqrt{49m^{2} + 4}$
$\Rightarrow 16 + 16m^{2} = 49m^{2} + 4$
$\Rightarrow 12 = 33m^{2}$
$\Rightarrow m^{2} = \frac{12}{33} \Rightarrow m = \frac{2}{\sqrt{11}}$
$\therefore y = \frac{2}{\sqrt{11}}x + 4\sqrt{1 + \frac{4}{11}}$
$= \frac{2}{\sqrt{11}}x + 4\sqrt{\frac{15}{11}}$