Question

Let complex numbers $\alpha$and $\frac{1}{\overline{\alpha }}$ 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 $z_0=x_0+i{y}_0$ satisfies the equation $2|z_0{|}^2=r^2+2$, then $|\alpha |=$

• A. $\frac{1}{\sqrt{2}}$
• B. $\frac{1}{2}$
• C. $\frac{1}{\sqrt{7}}$
• D. $\frac{1}{3}$

Answer 1

The correct option is C $\frac{1}{\sqrt{7}}$

$|\alpha -z_0|=r$..............(1)

$|\frac{1}{\overline{\alpha }}-z_0|=2r$........$\left(m\right)$

$2|z_0{|}^2=r^2+2$.......(3)

$\alpha .\overline{\alpha }=|\alpha {|}^2\Rightarrow \frac{1}{\overline{\alpha }}=\frac{\alpha }{|\alpha {|}^2}$

Putting $\frac{1}{\overline{\alpha }}=\frac{\alpha }{|\alpha {|}^2}$ in $\left(m\right)$, we get

$|\frac{\alpha }{|\alpha {|}^2}-z_0|=2r$........(2)

Squaring (1), we get,

$|\alpha {|}^2+|z_0{|}^2-(\alpha \overline{z_0}+\overline{\alpha }{z}_0)=r^2$

$\Rightarrow |\alpha {|}^2+\frac{r^2}{2}+1-(\alpha \overline{z_0}+\overline{\alpha }{z}_0)=r^2....\left(4\right)$

$\Rightarrow \frac{|\alpha {|}^2}{|\alpha {|}^4}+|z_0{|}^2-\left(\frac{\alpha \overline{z_0}+\overline{\alpha }{z}_0}{|\alpha {|}^2}\right)=4r^2$

$\Rightarrow \frac{1}{|\alpha {|}^2}(1-(\alpha \overline{z_0}+\overline{\alpha }{z}_0\left)\right)=4r^2-|z_0{|}^2$

$=4r^2-\left(\frac{r^2+2}{2}\right)$

$=\frac{7r^2}{2}-1$

$\Rightarrow 1-(\alpha \overline{z_0}+\overline{\alpha }{z}_0)=|\alpha {|}^2(\frac{7r^2}{2}-1)$

$\Rightarrow \alpha \overline{z_0}+\overline{\alpha }{z}_0=1+|\alpha {|}^2(1-\frac{7r^2}{2})$

Substituting in (4), we get

$|\alpha {|}^2+\frac{r^2}{2}+1-(1+|\alpha {|}^2(1-\frac{7r^2}{2}))=r^2$

$\Rightarrow \frac{r^2}{2}+|\alpha {|}^2\left(\frac{7r^2}{2}\right)=r^2$

$\Rightarrow |\alpha {|}^2\left(\frac{7r^2}{2}\right)=\frac{r^2}{2}$

$\Rightarrow |\alpha {|}^2=\frac{1}{7}$

$\therefore |\alpha |=\frac{1}{\sqrt{7}}$