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

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$\frac{1}{\sqrt{2}}$

$\frac{1}{\sqrt{7}}$

$\frac{1}{2}$

$\frac{1}{3}$

answer is C.

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Detailed Solution

$\begin{array}{l} α l i e s o n |z - z_{0}| = r \Rightarrow |α - z_{0}| = r \\ \frac{1}{\bar{α}} l i e s o n |z - z_{0}| = 2 r \Rightarrow |\frac{1}{\bar{α}} - z_{0}| = 2 r \\ ∴ z_{0} = 2 α - \frac{1}{α} \\ ∵ 2 {|z_{0}|}^{2} = r^{2} + 2 \\ ∴ 2 {|2 α - \frac{1}{α}|}^{2} = r^{2} + 2 \\ \Rightarrow 2 {|2 α - \frac{1}{α}|}^{2} = {|α - \frac{1}{α}|}^{2} + 2 \\ \Rightarrow 7 | α |^{2} + \frac{1}{| α |^{2}} - 8 = 0 \Rightarrow | α |^{2} = 1 or \frac{1}{7} \Rightarrow | α | = 1 or \frac{1}{\sqrt{7}} \end{array}$