Let the chord of contact is $PQ$ which touches $x^2+4y^2=4$ at $R$
Now, assume $R=\left(2\cos \theta ,2\sin \theta \right)$
Equation of the chord of contact $PQ$ is $hx+2yk=6\dots ..\left(\mathrm{i}\right)$
Again, the equation of the tangent $PQ$ is $\frac{x\cos \theta }{2}+\frac{y\sin \theta }{1}=1\dots ..\left(\mathrm{i}\mathrm{i}\right)$
From $\left(\mathrm{i}\right)$ and $\left(\mathrm{i}\mathrm{i}\right)$, we get,

 $\frac{h2}{\cos \theta }=\frac{2k}{sin\theta }=6$
$$\begin{gathered} \cos \theta =\frac{h}{3},\sin \theta =\frac{k}{3} \\ \Rightarrow x^2+y^2=9 \end{gathered}$$

Hence, the perimeter of the circle is $6\mathrm{\pi }$ units