Let complex numbers $\alpha$ and $\frac{1}{\alpha}$ lie on circles $(x - x_{0})^{2} + (y - y_{0})^{2} = r^{2}$ and $(x - x_{0})^{2} + (y - y_{0})^{2} = 4r^{2}$ respectively. If $z_{0} = x_{0} + iy_{0}$ satisfies the equation $2|z_{0}|^{2} = r^{2} + 2$, then $|\alpha| =$

$\alpha \text{ on } (x-x_0)^2+(y-y_0)^2=r^2$ $\dfrac{1}{\alpha} \text{ on } (x-x_0)^2+(y-y_0)^2=4r^2$ $z_0 = x_0 + iy_0, \quad 2|z_0|^2 = r^2 + 2$ $\text{To find: } |\alpha|$ $1.\ \left|\alpha - z_0\right|^2 = r^2, \quad \left|\dfrac{1}{\bar\alpha} - z_0\right|^2 = 4r^2$ $2.\ \text{multiply the second by } |\alpha|^2$ $3.\ \text{subtract and cancel the common term, then use } 2|z_0|^2 = r^2 + 2$ $|\alpha - z_0|^2 = r^2$ $|\alpha|^2 - 2\,\text{Re}(\alpha\bar z_0) + |z_0|^2 = r^2 \quad \cdots (i)$ $\left|\dfrac{1}{\bar\alpha} - z_0\right|^2 = 4r^2$ $\left|\dfrac{1}{\alpha} - \bar z_0\right|^2 = 4r^2$ $\left|1 - \alpha\bar z_0\right|^2 = 4r^2|\alpha|^2$ $1 - 2\,\text{Re}(\alpha\bar z_0) + |z_0|^2|\alpha|^2 = 4r^2|\alpha|^2 \quad \cdots (ii)$ $(ii) - (i): \quad 1 - |z_0|^2 + r^2 = |\alpha|^2\left(1 - |z_0|^2 + 4r^2\right)$ $2|z_0|^2 = r^2 + 2 \ \Rightarrow \ |z_0|^2 = \dfrac{r^2+2}{2}$ $1 - |z_0|^2 + r^2 = 1 - \dfrac{r^2+2}{2} + r^2 = \dfrac{r^2}{2}$ $1 - |z_0|^2 + 4r^2 = 1 - \dfrac{r^2+2}{2} + 4r^2 = \dfrac{7r^2}{2}$ $\dfrac{r^2}{2} = |\alpha|^2 \cdot \dfrac{7r^2}{2}$ $1 = 7|\alpha|^2$ $|\alpha|^2 = \dfrac{1}{7}$ $\boxed{|\alpha| = \dfrac{1}{\sqrt{7}}} \quad \Rightarrow \quad \text{option (3)}$