JEE Advanced 2013 · Paper 1, Mathematics question 41

Set-up. Let $a=|\alpha|^2$; note $\dfrac1{\bar\alpha}=\dfrac\alpha a$.
$$|\alpha-z_0|^2=r^2:\quad a-2\,\mathrm{Re}(\alpha\bar z_0)+|z_0|^2=r^2,$$
$$\left|\tfrac\alpha a-z_0\right|^2=4r^2:\quad\frac1a-\frac2a\mathrm{Re}(\alpha\bar z_0)+|z_0|^2=4r^2.$$

Eliminate. Multiply the second by $a$ and subtract from the first:
$$(a-1)\left(1-|z_0|^2\right)=r^2(1-4a).$$

Use the condition. $|z_0|^2=\frac{r^2+2}2$, so $1-|z_0|^2=-\frac{r^2}2$: $-\frac{a-1}2=1-4a$, giving $7a=1$.

So $|\alpha|=\dfrac1{\sqrt7}$.

(C)