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 $2{\left|z_0\right|}^2=r^2+2$, then $\left|\alpha \right|$ =

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

Answer 1

Answer: (C)

$1/\sqrt{7}$

The circles are $(x-x_0{)}^2+(y-y_0{)}^2=r^2$ and $(x-x_0{)}^2+(y-y_0{)}^2=4r^2$

The centre is $z_0=x_0+i{y}_0$

Hence circles can be written as $|z-z_0|=r$ and $|z-z_0|=2r$.

$\alpha$ and $\frac{1}{\overline{\alpha }}$ lies on these circles.

Hence $|\alpha -z_0|=r$ and $|\frac{1}{\overline{\alpha }}-z_0|=2r$

$⟹|\alpha -z_0{|}^2=r^2$ and $|\frac{1}{\overline{\alpha }}-z_0{|}^2=(2r{)}^2$

$|\alpha -z_0{|}^2=r^2$

$(\alpha -z_0)\left(\overline{\alpha -z_0}\right)=r^2$

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

$|\alpha {|}^2+|z_0{|}^2-z_0\overline{\alpha }-\alpha \overline{z_0}=r^2$

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

Similarly $|\frac{1}{\overline{\alpha }}-z_0{|}^2=(2r{)}^2=4r^2$

$|1-z_0\overline{\alpha }{|}^2=4r^2|\alpha {|}^2$

$(1-z_0\overline{\alpha })(1-\overline{z_0}\alpha )=4r^2|\alpha {|}^2$

$1-z_0\overline{\alpha }-\overline{z_0}\alpha +|\alpha {|}^2|z_0{|}^2=4r^2|\alpha {|}^2$

$\Rightarrow$ $z_0\overline{\alpha }+\alpha \overline{z_0}=1+|\alpha {|}^2|z_0{|}^2-4r^2|\alpha {|}^2$ ------(2)

From (1) and (2)

$|\alpha {|}^2+|z_0{|}^2-r^2=1+|\alpha {|}^2|z_0{|}^2-4r^2|\alpha {|}^2$

$|\alpha {|}^2+\frac{r^2+2}{2}-r^2=1+|\alpha {|}^2(\frac{r^2+2}{2})-4r^2|\alpha {|}^2$ (Given $2|z_0{|}^2=r^2+2$)

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

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