$S\equiv x^2+y^2-4=0, r_1=2, C_1(0,0)$

Let center of $S^{\text{'}}=0,$ is $C_2\left(h,k\right), r_2=\frac{5\sqrt{2}}{2}$

Equation of $S^{\text{'}}=0, {\left(x-h\right)}^2+{\left(y-k\right)}^2=\frac{25}{2}$

$\Rightarrow x^2+y^2-2xh-2yk+h^2+k^2-\frac{25}{2}=0$

Equation of $AB S^{\text{'}}-S=0$

$2xh+2yk=h^2+k^2-\frac{17}{2} ...\left(\mathrm{i}\right)$

Slope of $AB=\frac{-2h}{2k}=\frac{1}{4}\Rightarrow -4h=k ...\left(\mathrm{ii}\right)$

Now, length of $AB$ will be maximum if $AB$ is passing through $C_1(0,0)$

$\Rightarrow C_{1}C=0$

$\left|\frac{-h^2-k^2+{\displaystyle \frac{17}{2}}}{\sqrt{4h^2+4k^2}}\right|=0$

$\Rightarrow h^2+k^2=\frac{17}{2} ...\left(\mathrm{iii}\right)$

Using $\left(\mathrm{ii}\right)$ and $\left(\mathrm{iii}\right)$

 $17h^2=\frac{17}{2}=h=\pm \frac{1}{\sqrt{2}}=\pm \frac{\sqrt{2}}{2}$

$C_2\left(\pm \frac{\sqrt{2}}{2},\mp 2\sqrt{2}\right)$