Let the complex numbers alpha and left(frac{1 | vedclass

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(C) As point $\alpha$ lies on the circle $\left(x-x_0\right)^2+\left(y-y_0\right)^2=r^2$,we have $|\alpha-z_0|^2=r^2$,where $z_0=x_0+iy_0$.Expanding t

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

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(C) As point $\alpha$ lies on the circle $\left(x-x_0ight)^2+\left(y-y_0ight)^2=r^2$,we have $|\alpha-z_0|^2=r^2$,where $z_0=x_0+iy_0$.Expanding this,we get $|\alpha|^2+|z_0|^2-(\alpha\bar{z}_0+\bar{\alpha}z_0)=r^2 \quad \ldots (i)$Since $\frac{1}{\bar{\alpha}}$ lies on the circle $\left(x-x_0ight)^2+\left(y-y_0ight)^2=4r^2$,we have $|\frac{1}{\bar{\alpha}}-z_0|^2=4r^2$.Expanding this,we get $\frac{1}{|\alpha|^2}+|z_0|^2-(\frac{\alpha\bar{z}_0}{|\alpha|^2}+\frac{\bar{\alpha}z_0}{|\alpha|^2})=4r^2$.Multiplying by $|\alpha|^2$,we get $1+|z_0|^2|\alpha|^2-(\alpha\bar{z}_0+\bar{\alpha}z_0)=4r^2|\alpha|^2 \quad \ldots (ii)$Subtracting $(i)$ from $(ii)$,we get $(|\alpha|^2-1)|z_0|^2 - (|\alpha|^2-1) = r^2(4|\alpha|^2-1)$.$(|\alpha|^2-1)(|z_0|^2-1) = r^2(4|\alpha|^2-1)$.Given $2|z_0|^2=r^2+2$,we have $|z_0|^2-1 = \frac{r^2}{2}$.Substituting this,$(|\alpha|^2-1)\frac{r^2}{2} = r^2(4|\alpha|^2-1)$.Dividing by $r^2$ (assuming $r 
eq 0$),we get $\frac{|\alpha|^2-1}{2} = 4|\alpha|^2-1$.$|\alpha|^2-1 = 8|\alpha|^2-2$.$7|\alpha|^2=1 \Rightarrow |\alpha|=\frac{1}{\sqrt{7}}$.

Let the complex numbers $\alpha$ and $\left(\frac{1}{\bar{\alpha}}\right)$ lie 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 $z_0=x_0+i y_0$ satisfies the equation $2|z_0|^2=r^2+2$,then $|\alpha|=$

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