Let complex numbers $\alpha \text{and}\frac{1}{\alpha }$ lies on circle ${\left(x-x_0\right)}^2{\left(y-y_0\right)}^2=r^2\text{and}{\left(x-x_0\right)}^2+{\left(y-y_0\right)}^2=4r^2$ respectively. If $z_0=x_0+i{y}_0$ satisfies the equation $2{\left|z_0\right|}^2=r^2+2$ then $\left|\alpha \right|$ is equal to (a) $\frac{1}{\sqrt{2}}$ (b) $\frac{1}{2}$ (c) $\frac{1}{\sqrt{7}}$ (d) $\frac{1}{3}$

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