Answer:
Option C
Explanation:
Concept involved intersection of circles , the basic concept is to equations simulataneously and using properties of modules of complex numbers.
Formula used |z|2=$z.\overline{z}$
and $|z_{1}-z_{2}|^{2}=(z_{1}-z_{2})(\overline{z_{1}}-\overline{z_{2}})$
= $|z_{1}|^{2}-z_{1}\overline{z_{2}}-z_{2}\overline{z_{1}}+|z_{2}|^{2}$
Here,
$(x-x_{0})^{2}+(y-y_{0})^{2}=r^{2} $ and
$(x-x_{0})^{2}+(y-y_{0})^{2}=4r^{2} $ could be written as,
$|z-z_{0}|^{2}=r^{2}$ and $ |z-z_{0}|^{2}=4r^{2}$
since ,$\alpha$ and $\frac{1}{\overline{\alpha}}$ lies on first and second respectively,
$\therefore$ $|\alpha-z_{0}|^{2}=r^{2}$ and$ |\frac{1}{\overline{\alpha}}-z_{0}|^{2}=4r^{2}$
$\Rightarrow$ $(\alpha-z_{0})(\overline{\alpha}-\overline{z_{0}})=r^{2}$
$\Rightarrow$ $|\alpha|^{2}-z_{0}\overline{\alpha}-\overline{z_{0}}\alpha+|z_{0}|^{2}=r^{2}$ .....(i)
and $|\frac{1}{\overline{\alpha}}-z_{0}|^{2}=4r^{2}$
$\Rightarrow$ $(\frac{1}{\overline{\alpha}}-z_{0})(\frac{1}{\alpha}-\overline{z_{0}})=4r^{2}$
$\Rightarrow$ $\frac{1}{|\alpha|^{2}}-\frac{z_{0}}{\alpha}-\frac{\overline{z_{0}}}{\overline{\alpha}}+|z_{0}|^{2}=4r^{2}$
Since $|\alpha|^{2}=\alpha.\alpha$
$\Rightarrow$ $\frac{1}{|\alpha|^{2}}-\frac{z_{0}.\overline{\alpha}}{|\alpha|^{2}}-\frac{\overline{z_{0}}}{|{\alpha}|^{2}}\alpha+|z_{0}|^{2}=4r^{2}$
$\Rightarrow$ $1-z_{0}\overline{\alpha}-\overline{z_{0}}\alpha+|\alpha|^{2}|z_{0}|^{2}=4r^{2}|\alpha|^{2}$ ...(ii)
On substracting Eqs (i) and (ii) , we get
$(|\alpha|^{2}-1)+|z_{0}|^{2}(1-|\alpha|^{2})$
= $r^{2}(1-4|\alpha|^{2})$
$\Rightarrow$ $(|\alpha|^{2}-1)(1-|z_{0}|^{2})=r^{2}(1-4|\alpha|^{2})$
$\Rightarrow$ $(|\alpha|^{2}-1)\left(1-\frac{r^{2}+2}{2}\right)$
$=r^{2}(1-4|\alpha|^{2})$
Given, $|z_{0}|^{2}=\frac{r^{2}+2}{2}$
$\Rightarrow$ $(|\alpha|^{2}-1).\left(\frac{-r^{2}}{2}\right)=r^{2}(1-4|\alpha|^{2})$
$\Rightarrow$ $|\alpha|^{2}-1=-2+8|\alpha|^{2}$
$\Rightarrow$ $7|\alpha|^{2}=1$
$\therefore$ $|\alpha|=\frac{1}{\sqrt{7}}$
