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 $\left(x-x_{0}\right)^{2}+\left(y-y_{0}\right)^{2}=4 r^{2}$ 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 $\left(x-x_{0}\right)^{2}+\left(y-y_{0}\right)^{2}=4 r^{2}$
$\therefore \left|\frac{1}{\alpha}-z_{0}\right|^{2}=4 r^{2}$
$\Rightarrow \frac{1}{|\alpha|^{2}}+\left |z_{0}\right|^{2}-\left(\frac{\bar{\alpha}_{0}}{|\alpha|^{2}}+\frac{\bar{\alpha} z_{0}}{\left.|\alpha|\right|^{2}}\right)=4 r^{2}$
$\Rightarrow 1+\left |z_{0}\right|^{2}|\alpha|^{2}-\left(\alpha \bar{z}_{0}+\bar{\alpha} z_{0}\right)=4 r^{2}|\alpha|^{2}$ ...(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
$\left(|\alpha|^{2}-1\right) \frac{r^{2}}{2}=r^{2}\left(4|\alpha|^{2}-1\right)$
$\Rightarrow |\alpha|^{2}-1=8|\alpha|^{2}-2$
$\Rightarrow 7|\alpha|^{2}=1$
$\Rightarrow |\alpha|=\frac{1}{\sqrt{7}}$