$\mathrm{\alpha }$ lie on (x – x₀)² + (y – y₀)² = r²

 $\therefore$a = z₀ + re^{i$\mathrm{\theta }$}               …….(1)

Similarly,  $\frac{1}{\mathrm{\alpha }}$  lie on (x – x₀)² + (y – y₀)² = 4r²

 $\therefore \frac{1}{\mathrm{\alpha }}={\mathrm{z}}_0+2{\mathrm{re}}^{\mathrm{i\theta }}$               …….(2)

From (2) – (1)

 $\therefore \frac{1}{\mathrm{\alpha }}-\mathrm{\alpha }={\mathrm{e}}^{\mathrm{i\theta }}\Rightarrow {\mathrm{r}}^2(\frac{1}{\overline{\mathrm{\alpha }}}-\mathrm{\alpha })(\frac{1}{\mathrm{\alpha }}-\overline{\mathrm{\alpha }})=\frac{1}{|\mathrm{\alpha }{|}^2}+|\mathrm{\alpha }{|}^2=2|{\mathrm{z}}_0{|}^2$     …….(3)

From (1) and (2) again.

 $\frac{\mathrm{\alpha }-{\mathrm{z}}_0}{\frac{1}{\overline{\mathrm{\alpha }}}-{\mathrm{z}}_0}=\frac{1}{2}$

 $\therefore {\mathrm{z}}_0=2\mathrm{\alpha }-\frac{1}{\mathrm{\alpha }}\Rightarrow |{\mathrm{z}}_0{|}^2=(2\mathrm{\alpha }-\frac{1}{\overline{\mathrm{\alpha }}})(2\overline{\mathrm{\alpha }}-\frac{1}{\mathrm{\alpha }})$

2 | z₀ |2 = 8 | $\mathrm{\alpha }$ |2 – 8 +  $\frac{2}{|\mathrm{\alpha }{|}^2}$                   ……(4)

From (3) and (iv)

7 | $\mathrm{\alpha }$ |4 – 8 | $\mathrm{\alpha }$ |2 + 1 = 0

 $\therefore$     | $\mathrm{\alpha }$ |2 = $\frac{1}{7}$  + 1

 $\therefore$      | $\mathrm{\alpha }$ | =  $\frac{1}{\sqrt{7}}$ .