Let complex numbers alpha and frac{1}{bar{alp | vedclass

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

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(C) The given circles are $|z-z_0|=r$ and $|z-z_0|=2r$.Since $\alpha$ lies on the first circle,$|\alpha-z_0|=r$,which implies $|\alpha-z_0|^2=r^2$.Expanding this,$|\alpha|^2 - z_0\bar{\alpha} - \bar{z}_0\alpha + |z_0|^2 = r^2$  $(1)$.Since $\frac{1}{\bar{\alpha}}$ lies on the second circle,$|\frac{1}{\bar{\alpha}}-z_0|=2r$,which implies $|\frac{1}{\bar{\alpha}}-z_0|^2=4r^2$.Using $\bar{\alpha}\alpha = |\alpha|^2$,we have $|\frac{\alpha}{|\alpha|^2}-z_0|^2=4r^2$.Expanding this,$\frac{1}{|\alpha|^2} - \frac{z_0\bar{\alpha}}{|\alpha|^2} - \frac{\bar{z}_0\alpha}{|\alpha|^2} + |z_0|^2 = 4r^2$.Multiplying by $|\alpha|^2$,$1 - z_0\bar{\alpha} - \bar{z}_0\alpha + |z_0|^2|\alpha|^2 = 4r^2|\alpha|^2$  $(2)$.Subtracting $(1)$ from $(2)$,$(|z_0|^2|\alpha|^2 - |z_0|^2) + (1 - |\alpha|^2) = 4r^2|\alpha|^2 - r^2$.$|z_0|^2(|\alpha|^2-1) - (|\alpha|^2-1) = r^2(4|\alpha|^2-1)$.$(|z_0|^2-1)(|\alpha|^2-1) = r^2(4|\alpha|^2-1)$.Given $2|z_0|^2 = r^2+2$,so $|z_0|^2 = \frac{r^2+2}{2}$.Substituting this,$(\frac{r^2+2}{2}-1)(|\alpha|^2-1) = r^2(4|\alpha|^2-1)$.$\frac{r^2}{2}(|\alpha|^2-1) = r^2(4|\alpha|^2-1)$.$\frac{1}{2}|\alpha|^2 - \frac{1}{2} = 4|\alpha|^2 - 1$.$\frac{1}{2} = \frac{7}{2}|\alpha|^2$.$|\alpha|^2 = \frac{1}{7} \Rightarrow |\alpha| = \frac{1}{\sqrt{7}}$.

Let complex numbers $\alpha$ and $\frac{1}{\bar{\alpha}}$ lie on the 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 $|\alpha|=$

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