Let complex numbers α and $\frac{1}{\overline{\alpha }}$   lie on circles ${\left(x-x_0\right)}^2+{\left(y-y_0\right)}^2={\text{r}}^2\text{, }$  and  ${\left(x-x_0\right)}^2+{\left(y-y_0\right)}^2={4\text{r}}^2\text{, }$respectively. If $z_0=x_0+i{y}_0$ satisfies the equation $2{\left|z_0\right|}^2=r^2+2\text{,}\text{then}\left|\alpha \right|=$

\(\left| z - z _0\right|= r\)
\(\left|z-z_0\right|=2 r\)
\(\left|\alpha- z _0\right|= r\)
\(\left|\frac{1}{\bar{\alpha}}- z _0\right|=2 r \quad \alpha \bar{\alpha}=|\alpha|^2\)
\(\left|\frac{\alpha}{|\alpha|^2}-z_0\right|=2 r\)
\(\left(\alpha- z _0\right)\left(\bar{\alpha}-\overline{ z }_0\right)= r ^2 \Rightarrow|\alpha|^2\) \(-~ z _0 \bar{\alpha}-\alpha \overline{ z }_0+\left| z _0\right|^2= r ^2\)
\(\left(\frac{\alpha}{|\alpha|^2}- z _0\right)\left(\frac{\bar{\alpha}}{|\alpha|^2}-\bar{z}_0\right)=4 r ^2 \Rightarrow\) \(\frac{|\alpha|^2}{|\alpha|^4}-\frac{\bar{z}_0 \alpha}{|\alpha|^2}-\frac{\bar{z}_0 \alpha}{|\alpha|^2}+\left| z _0\right|^2=4 r ^2\)
\(1- z _0 \bar{\alpha}-\bar{z}_0 \alpha+\left| z _0\right|^2|\alpha|^2=4 r ^2|\alpha|^2 \)
\( \Rightarrow\left(\left|\alpha^2\right|-1\right)+\left| z _0\right|^2\left(1-|\alpha|^2\right)= r ^2\left(1-4 \alpha^2\right) \)
\( \left(|\alpha|^2-1\right)\left(1-\frac{ r ^2+2}{2}\right)= r ^2\left(1-4|\alpha|^2\right) \)
\( \left(|\alpha|^2-1\right)\left(\frac{- r ^2}{2}\right)= r ^2\left(1-4|\alpha|^2\right) \)
\( |\alpha|^2-1=-2+8|\alpha|^2 \)
\( 1=7|\alpha|^2 \quad \Rightarrow \quad|\alpha|=\frac{1}{\sqrt{7}}\)