The length of the portion of the common tangent to x^ 2 + y^ 2 = 16...

Equation of the ellipse is $\dfrac{x^{2}}{25} + \dfrac{y^{2}}{9} = 1$
Any point on the ellipse is $\left(5cos \theta,3sin \theta\right)$ at which the equation of the tangent is
$\dfrac{x}{5}cos \theta + \dfrac{y}{3}sin \theta = 1$
If it is also a tangent to the circle $x^{2} + y^{2} = 16$, then its distance~ from the origin is equal to the radius,&
$⇒ \dfrac{1}{\left(\dfrac{cos^{2}\theta}{25} + \dfrac{sin^{2}\theta}{9}\right)} = 4$
$⇒ \dfrac{\cos^{2}\theta}{25} + \dfrac{1 - \cos^{2}\theta}{9} = \dfrac{1}{16}$
$⇒ cos \theta = \dfrac{5\sqrt{7}}{16}$
If $l$ be the length of the tangent then $l^{2} = S^{'}$
or $l^{2} = 25cos^{2}\theta + 9sin^{2}\theta - 16 = 16cos^{2}\theta - 7$
$= 16 . \dfrac{25.7}{{\left(16\right)}^{2}} - 7 = \dfrac{175 - 112}{16} = \dfrac{63}{16}$
$⇒ l = \dfrac{3}{4}\sqrt{7}$& units