Mathematics - Complex Number Question with Solution | TestHub

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