If a circle $\left(x-1\right){}^2+y^2=r^2$ touches the ellipse $x^2+4y^2=16$ internally, then $r=$

Given circle is ${\left(x-1\right)}^2+y^2=r^2....\left(i\right)$

Here, centre of circle is $\left(1,0\right) \&$radius$=r$.

Given ellipse is  $x^2+4y^2=16$ i.e. $\frac{x^2}{16}+\frac{y^2}{4}=1$

Comparing with $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$a=4, b=2$

Also given that the circle touches the ellipse internally as shown in figure.

Let any point on ellipse be  $P\left(4\cos \theta , 2\sin \theta \right)$

$\therefore$Equation of normal of ellipse

$ax.\sec \theta -by.\mathrm{cosec}\theta =a^2-b^2$

$\therefore 4x.\sec \theta -2y.\mathrm{cosec}\theta =12$

And it will pass through the centre of the circle $\left(1,0\right)$

$\therefore 4\times 1.\sec \theta -0=12$

$\Rightarrow \sec \theta =3$

$\Rightarrow \cos \theta =\frac{1}{3}$

And $\sin \theta =\sqrt{1-\frac{1}{9}}=\frac{2\sqrt{2}}{3}$

Now, the point is  $\left(4\cos \theta , 2\sin \theta \right)\equiv \left(\frac{4}{3}, \frac{4\sqrt{2}}{3}\right)$

Radius $r=$Distance between $\left(1,0\right) \& \left(\frac{4}{3},\frac{2\sqrt{2}}{3}\right)$

 $=\sqrt{{\left(1-\frac{4}{3}\right)}^2+{\left(0-\frac{4\sqrt{2}}{3}\right)}^2}$

$=\sqrt{\frac{1}{9}+\frac{32}{9}}$

$=\sqrt{\frac{11}{3}}$