Question

Let the complex numbers $\alpha$ and $\left(\frac{1}{\alpha }\right)$ lie on circles ${\left(x-x_0\right)}^2+{\left(y-y_0\right)}^2=r^2$ and ${\left(x-x_0\right)}^2+{\left(y-y_0\right)}^2=4r^2$ respectively. If $z_0+i{y}_0$ satisfies the equation $2{\left|z_0\right|}^2=r^2+2$ then $|\alpha |=$

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

Answer 1

Answer: (C)

As point $\alpha$ lies on the circle
${\left(x-x_0\right)}^2+{\left(y-y_0\right)}^2=r^2$
$\therefore {\left|\alpha -z_0\right|}^2=r^2,$ where $z_0=x_0+i{y}_0$
$\Rightarrow |\alpha {|}^2+{\left|z_0\right|}^2-\left(\alpha {\bar{z}}_0+\bar{\alpha }{z}_0\right)=r^2$ ...(i)
$\because \frac{1}{\alpha }$ lies on the circle ${\left(x-x_0\right)}^2+{\left(y-y_0\right)}^2=4r^2$
$\therefore {\left|\frac{1}{\alpha }-z_0\right|}^2=4r^2$
$\Rightarrow \frac{1}{|\alpha {|}^2}+{\left|z_0\right|}^2-\left(\frac{{\bar{\alpha }}_0}{|\alpha {|}^2}+\frac{\bar{\alpha }{z}_0}{{|\alpha ||}^2}\right)=4r^2$
$\Rightarrow 1+{\left|z_0\right|}^2|\alpha {|}^2-\left(\alpha {\bar{z}}_0+\bar{\alpha }{z}_0\right)=4r^2|\alpha {|}^2$ ...(ii)
By subtracting Eqs. (i) and (ii), we get
$1-|\alpha {|}^2-{\left|z_0\right|}^2\left(1-|\alpha {|}^2\right)=r^2\left(4|\alpha {|}^2-1\right)$
$\Rightarrow \left(|\alpha {|}^2-1\right)\left({\left|z_0\right|}^2-1\right)=r^2\left(4|\alpha {|}^2-1\right)$
$\because {\left|z_0\right|}^2=\frac{r^2+2}{2},$ we get
$\left(|\alpha {|}^2-1\right)\frac{r^2}{2}=r^2\left(4|\alpha {|}^2-1\right)$
$\Rightarrow |\alpha {|}^2-1=8|\alpha {|}^2-2$
$\Rightarrow 7|\alpha {|}^2=1$
$\Rightarrow |\alpha |=\frac{1}{\sqrt{7}}$