JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · Complex Numbers

Let $\alpha$ lie on $|z-z_0|=r$, so $|\alpha -z_0{|}^2=r^2$. Since $\frac{1}{\bar{\alpha }}$ lies on $|z-z_0|=2r$, we have $\left|\frac{1}{\bar{\alpha }}-z_0\right|=2r$, which gives $|1-\bar{\alpha }{z}_0|=2r|\alpha |$, hence $|1-\bar{\alpha }{z}_0{|}^2=4r^2|\alpha {|}^2$. Subtracting the first equation from this: $|1-\bar{\alpha }{z}_0{|}^2-|\alpha -z_0{|}^2=r^2(4|\alpha {|}^2-1)$. Expanding: $1+|\alpha {|}^2|z_0{|}^2-|\alpha {|}^2-|z_0{|}^2=r^2(4|\alpha {|}^2-1)$, so $(1-|\alpha {|}^2)(1-|z_0{|}^2)=r^2(4|\alpha {|}^2-1)$. Using $2|z_0{|}^2=r^2+2$, we get $1-|z_0{|}^2=-\frac{r^2}{2}$. Substituting: $(1-|\alpha {|}^2)\left(-\frac{r^2}{2}\right)=r^2(4|\alpha {|}^2-1)$, which simplifies to $(1-|\alpha {|}^2)(7|\alpha {|}^2-1)=0$. Since $|\alpha |\ne 1$, we have $|\alpha {|}^2=\frac{1}{7}$, thus $|\alpha |=\frac{1}{\sqrt{7}}$.