Let the complex numbers α and 1 ¯ α lie on the circles |z - z _ 0 |...

Given: ${\left|z - {z}_{0}\right|}^{2} = 4$

$⇒ \left(z - {z}_{0}\right)\left(z - {z}_{0}\right) = 4$

$⇒ z z - {z}_{0}z - {z}_{0} z + {\left|{z}_{0}\right|}^{2} = 4$

$⇒ {\left|z\right|}^{2} - {z}_{0}z - {z}_{0} z = 2\left\{\text{as}{\left|{z}_{0}\right|}^{2} = 2\right\}$

It is given that, $\alpha$ lies on ${\left|z - {z}_{0}\right|}^{2} = 4$.

$⇒ {\left|\alpha\right|}^{2} - \alpha{z}_{0} - {z}_{0} \alpha - 2 = 0 . . . \left(i\right)$

We know that, $\alpha \cdot \alpha = {\left|\alpha\right|}^{2} = {\left|\alpha\right|}^{2}$

${\left|z - {z}_{0}\right|}^{2} = 16$

$⇒ \left(1\overset{¯}{\alpha} - {z}_{0}\right)\left(\dfrac{1}{\alpha} - \overset{¯}{z}0\right) = 16$

$⇒ \left(1 - \overset{¯}{\alpha}{z}_{0}\right)\left(1 - \alpha\overset{¯}{z}0\right) = 16 | \alpha{|}^{2}$

$⇒ 1 - \overset{¯}{\alpha}{z}_{0} - \alpha\overset{¯}{z}0 + | \alpha{|}^{2}{\left|{z}_{0}\right|}^{2} = 16 | \alpha{|}^{2}$

$⇒ 1 - \overset{¯}{\alpha}{z}_{0} - \alpha\overset{¯}{z}0 = 14 | \alpha{|}^{2} . . . \left(ii\right)$

Using, \left(i\right)\text{and}\left(ii\right)

$⇒ 5 | \alpha{|}^{2} = 1$

$⇒ 100 | \alpha{|}^{2} = 20$