Let the Complex numbers  $\alpha$  and  $\frac{1}{\overline{\alpha }}$  lie on the circles ${|Z-Z_0|}^2=4$  and ${|Z-Z_0|}^2=16$  respectively where  $Z_0=1+i$. Then, The Value of  $100{\left|\alpha \right|}^2$  is 

    ${|Z-Z_0|}^2=4 Z=1+i\Rightarrow \left|Z\right|=\sqrt{2}$  
 $\begin{array}{l}\Rightarrow {|\alpha -Z_0|}^2=4\\ \Rightarrow (\alpha -Z_0)(\overline{\alpha }-{\overline{Z}}_0)=4\\ \Rightarrow {\left|\alpha \right|}^2+2-\alpha {\overline{Z}}_0-\overline{\alpha }{\text{ Z}}_0=4\\ {\left|\alpha \right|}^2-\alpha {\overline{Z}}_0-\overline{\alpha }{\text{ Z}}_0=\mathrm{2.............}\left(1\right)\end{array}$
 $\begin{array}{l}{|Z-Z_0|}^2=16\\ \Rightarrow {|\frac{1}{\overline{\alpha }}-Z_0|}^2=16\end{array}$
 $\begin{array}{l}\Rightarrow (\frac{1}{\overline{\alpha }}-Z_0)(\frac{1}{\alpha }-{\overline{Z}}_0)=16\\ (1-\overline{\alpha }{Z}_0)(1-\alpha {\overline{Z}}_0)=16\alpha \overline{\alpha }\end{array}$
 $1+\alpha \overline{\alpha }{Z}_0{\overline{Z}}_0-\overline{\alpha }{Z}_0-\alpha {\overline{Z}}_0=16\alpha \overline{\alpha }$
       $1+{\left|\alpha \right|}^{2}2-\overline{\alpha }{Z}_0-\alpha {\overline{Z}}_0=16{\left|\alpha \right|}^2$           $(\because {\left|Z\right|}^2=2)$
$1+\overline{\alpha }{Z}_0-\alpha {\overline{Z}}_0=14{\left|\alpha \right|}^2$                ………………(2)
(2) – (1) 
 $\begin{array}{l}\Rightarrow 1-{\left|\alpha \right|}^2=14{\left|\alpha \right|}^2-2\\ 15{\left|\alpha \right|}^2=3\Rightarrow {\left|\alpha \right|}^2=\frac{1}{5}\\ 100{\left|\alpha \right|}^2=20\end{array}$