Let complex numbers $\alpha$ and $\dfrac{1}{\bar{\alpha}}$ lie on 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 $|\alpha|$ is equal to
Step-by-Step Solution
Key Concept: The key is eliminating $\text{Re}(\alpha\bar{z}_0)$ between the two expanded circle equations, then substituting $2|z_0|^2=r^2+2$ to get a pure equation in $|\alpha|$.
**Step 1: Translate to modulus equations**
$|\alpha-z_0|=r$ ...(1) and $\left|\dfrac{1}{\bar{\alpha}}-z_0\right|=2r$ ...(2).
**Step 2: Expand (1)**
$|\alpha|^2-2\,\text{Re}(\alpha\bar{z}_0)+|z_0|^2=r^2$, so $2\,\text{Re}(\alpha\bar{z}_0)=|\alpha|^2+|z_0|^2-r^2$.
**Step 3: Expand (2) and multiply by $|\alpha|^2$**
From $|1/\bar{\alpha}-z_0|^2=4r^2$, multiplying through by $|\alpha|^2$: $1-2\,\text{Re}(\alpha\bar{z}_0)+|\alpha|^2|z_0|^2=4r^2|\alpha|^2$.
**Step 4: Eliminate cross term and substitute $r^2=2|z_0|^2-2$**
Substituting $2\,\text{Re}(\alpha\bar{z}_0)$ from step 2 and $r^2=2|z_0|^2-2$ into step 3, after simplification: $(|z_0|^2-1)(1+|\alpha|^2)=8|\alpha|^2(|z_0|^2-1)$. Dividing (assuming $|z_0|^2\neq1$): $1+|\alpha|^2=8|\alpha|^2 \Rightarrow 7|\alpha|^2=1 \Rightarrow |\alpha|=\dfrac{1}{\sqrt{7}}$.
Correct Answer: 3