Let complex numbers α and 1 ¯ α lie on circles (x - x _ 0 ) ^...

$\left|\text{z} - {\text{z}}_{0}\right| = \text{r}$
\left|\text{z} - {\text{z}}_{0}\right| = 2 \text{r}$<br />\left|\alpha - {\text{z}}_{0}\right| = \text{r}$
\left|1\overset{¯}{\alpha} - {\text{z}}_{0}\right| = 2\text{r}\alpha\overset{¯}{\alpha} = {\left|\alpha\right|}^{2}$<br />\left|\dfrac{\alpha}{{\left|\alpha\right|}^{2}} - {\text{z}}_{0}\right| = 2 \text{r}$
\left(\alpha - {\text{z}}_{0}\right) \left(\overset{¯}{\alpha} - {\overset{¯}{\text{z}}}_{0}\right) = {\text{r}}^{2} ⇒ {\left|\alpha\right|}^{2} - {\text{z}}_{0} \overset{¯}{\alpha} - \alpha {\overset{¯}{\text{z}}}_{0} + {\left|{\text{z}}_{0}\right|}^{2} = {\text{r}}^{2}$<br />\left(\dfrac{\alpha}{{\left|\alpha\right|}^{2}} - {\text{z}}_{0}\right) \left(\dfrac{\overset{¯}{\alpha}}{{\left|\alpha\right|}^{2}} - {\overset{¯}{\text{z}}}_{0}\right) = 4 {\text{r}}^{2} ⇒ \dfrac{{\left|\alpha\right|}^{2}}{{\left|\alpha\right|}^{4}} - \dfrac{{\overset{¯}{\text{z}}}_{0} \alpha}{{\left|\alpha\right|}^{2}} - \dfrac{{\overset{¯}{\text{z}}}_{0} \alpha}{{\left|\alpha\right|}^{2}} + {\left|{\text{z}}_{0}\right|}^{2} = 4 {\text{r}}^{2}$
$1 - {\text{z}}_{0} \overset{¯}{\alpha} - {\overset{¯}{\text{z}}}_{0} \alpha + {\left|{\text{z}}_{0}\right|}^{2} {\left|\alpha\right|}^{2} = 4 {\text{r}}^{2} {\left|\alpha\right|}^{2}$
$⇒ \left(\left|{\alpha}^{2}\right| - 1\right) + {\left|{\text{z}}_{0}\right|}^{2} \left(1 - {\left|\alpha\right|}^{2}\right) = {\text{r}}^{2} \left(1 - 4 {\alpha}^{2}\right)$
\left({\left|\alpha\right|}^{2} - 1\right) \left(1 - \dfrac{{\text{r}}^{2} + 2}{2}\right) = {\text{r}}^{2} \left(1 - 4 {\left|\alpha\right|}^{2}\right)$<br />\left({\left|\alpha\right|}^{2} - 1\right) \left(\dfrac{- {\text{r}}^{2}}{2}\right) = {\text{r}}^{2} \left(1 - 4 {\left|\alpha\right|}^{2}\right)$
${\left|\alpha\right|}^{2} - 1 = - 2 + 8 {\left|\alpha\right|}^{2}$
$1 = 7 {\left|\alpha\right|}^{2} ⇒ \left|\alpha\right| = \dfrac{1}{\sqrt{7}}$