If $$z=x+i y$$, then $$\left(x-x_0\right)^2+\left(y-y_0\right)^2=r^2$$ and $$\left(x-x_0\right)^2+\left(y-y_0\right)^2=4 r^2$$ can be written as $$\left|Z-Z_0\right|^2=r^2$$ and $$\left|Z-Z_0\right|^2=4 r^2$$ respectively where, $$\mathrm{Z}_0=x_0+i y_0$$ (given)

$$\Rightarrow\left|Z-Z_0\right|^2=r^2$$ and $$\left|Z-Z_0\right|^2=4 r^2$$

$$\Rightarrow\left(Z-Z_0\right)\left(\bar{Z}-\bar{Z}_0\right)=r^2$$ and $$\left(Z-Z_0\right)\left(\bar{Z}-\bar{Z}_0\right)=4 r^2\left(|Z|^2=Z \bar{Z}\right)$$

$$\begin{aligned} & \text { Given, } \alpha \text { and } \frac{1}{\bar{\alpha}} \text { line on circles }\left(x-x_0\right)^2+\left(y-y_0\right)^2 \\ & =r^2 \text { and }\left(x-x_0\right)^2+\left(y-y_0\right)^2=4 r^2 \text { respectively } \\ & \begin{array}{l} \therefore \quad\left(\alpha-Z_0\right)\left(\bar{\alpha}-\bar{Z}_0\right)=r^2 \text { and } \\ \quad\left(\frac{1}{\bar{\alpha}}-Z_0\right)\left(\frac{1}{\alpha}-\bar{Z}_0\right)=4 r^2 \end{array} \end{aligned}$$

$$\begin{aligned} \Rightarrow & |\alpha|^2-\alpha \bar{Z}_0-\bar{\alpha} Z_0+\left|Z_0\right|^2=r^2 \text { and } \\ & \frac{1}{|\alpha|^2}-\frac{1}{\bar{\alpha}} \bar{Z}_0-\frac{1}{\alpha} Z_0+\left|\bar{Z}_0\right|^2=4 r^2 \\ \Rightarrow & \alpha Z_0+\bar{\alpha} Z_0=|\alpha|^2+\left|Z_0\right|^2-r^2 \text { and } \\ & \frac{\alpha \bar{Z}_0}{|\alpha|^2}+\frac{\bar{\alpha} Z_0}{|\alpha|^2}=\frac{1}{|\alpha|^2}+\left|Z_0\right|^2-4 r^2 \\ \Rightarrow & \alpha \bar{Z}_0+\bar{\alpha} Z_0=|\alpha|^2+\left|Z_0\right|^2-r^2 =|\alpha|^2\left(\frac{1}{\left|\alpha^2\right|}+\left|Z_0^2\right|-4 r^2\right) \end{aligned}$$

$$\Rightarrow|\alpha|^2+\left|Z_0\right|^2-r^2=1+|\alpha|^2\left|Z_0\right|^2-4|\alpha|^2 r^2 \quad\text{... (i)}$$

Given, $$2\left|Z_0\right|^2=r^2+2$$

Put $$\left|Z_0\right|^2=\frac{r^2}{2}+1$$ in the equation (i)

$$\begin{aligned} & \Rightarrow|\alpha|^2+\frac{r^2}{2}+1-r^2=1+|\alpha|^2\left(\frac{r^2}{2}+1\right)-4|\alpha|^2 r^2 \\ & \Rightarrow|\alpha|^2-\frac{r^2}{2}=\frac{1}{2}|\alpha|^2 r^2+|\alpha|^2-4|\alpha|^2 r^2 \end{aligned}$$

$$\begin{array}{ll} \Rightarrow & -\frac{r^2}{2}=\frac{1}{2}|\alpha|^2 r^2-4|\alpha|^2 r^2 \\ \Rightarrow & -\frac{1}{2}=\frac{1}{2}(\alpha)^2-4|\alpha|^2 \\ \Rightarrow & \frac{-1}{2}=-\frac{7}{2}|\alpha|^2 \\ \Rightarrow & |\alpha|^2=\frac{1}{7} \\ \Rightarrow & |\alpha|=\frac{1}{\sqrt{7}} \end{array}$$

Hints :

(i) Apply the property $$Z \cdot \bar{Z}=|\bar{Z}|^2$$

(ii) If $$Z=(x+i y)$$ and $$Z_0=\left(x_0+i y_0\right)$$ then $$\left(x-x_0\right)^2+\left(y-y_0\right)^2=r^2$$ and $$\left(x-x_0\right)^2+ \left(y-y_0\right)^2=4 r^2$$ can be written as $$\left|Z-Z_0\right|^2=r^2$$ and $$\left|Z-Z_0\right|^2=4 r^2$$ respectively.