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 $\frac{1}{\sqrt{7}}$

$(x-x_0{)}^2+(y-y_0{)}^2=r^2$
$\Rightarrow |z-z_0|=r$
Point $\alpha$ lies on it.
$\Rightarrow |\alpha -z_0|=r$
$\Rightarrow |\alpha -z_0{|}^2=r^2\cdots \left(1\right)$

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

Subtracting $\left(1\right)$ from $\left(2\right)$,
${|1-\overline{\alpha }{z}_0|}^2-|\alpha -z_0{|}^2=r^2(4|\alpha {|}^2-1)$
$\Rightarrow 1+|\alpha {|}^2|z_0{|}^2-z_0\overline{\alpha }-\alpha {\overline{z}}_0-|\alpha {|}^2-|z_0{|}^2+z_0\overline{\alpha }+\alpha {\overline{z}}_0=r^2(4|\alpha {|}^2-1)$
$\Rightarrow 1+|\alpha {|}^2|z_0{|}^2-|\alpha {|}^2-|z_0{|}^2=r^2(4|\alpha {|}^2-1)$
$\Rightarrow (1-|\alpha {|}^2)(1-|z_0{|}^2)-r^2(4|\alpha {|}^2-1)=0$

$\because 2|z_0{|}^2=r^2+2\Rightarrow -r^2=2(1-|z_0{|}^2)$
$\Rightarrow (1-|\alpha {|}^2)(1-|z_0{|}^2)+2(1-|z_0{|}^2)(4|\alpha {|}^2-1)=0$
$\Rightarrow (1-|z_0{|}^2)(1-|\alpha {|}^2+8|\alpha {|}^2-2)=0$
$\Rightarrow (1-|z_0{|}^2)(7|\alpha {|}^2-1)=0$

$\Rightarrow (7|\alpha {|}^2-1)=0,(\because |z_0|>1)\therefore |\alpha |=\frac{1}{\sqrt{7}}$