Let complex numbers $\alpha$ and $\frac{1}{\bar{\alpha}}$ lies on circles $\left(x-x_0\right)^2+\left(y-y_0\right)^2=r^2$ and $\left(x-x_0\right)^2+\left(y-y_0\right)^2=4 r^2$, respectively. If $\mathrm{z}_0=\mathrm{x}_0+\mathrm{iy}$ satisfies the equation $2\left|\mathrm{z}_0\right|^2=\mathrm{r}^2+2$, then $|\alpha|=$