Discussion Forum

1)

Let complex numbers $\alpha$  and $\frac{1}{\alpha}$ lies 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 z₀=x₀+iy₀ satisfies the equation   2|z₀|²=r²+2 , then |$\alpha$|  is equal to

 

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,     $|z_{0}|^{2}=\frac{r^{2}+2}{2}$

$\Rightarrow$        $(|\alpha|^{2}-1).\left(\frac{-r^{2}}{2}\right)=r^{2}(1-4|\alpha|^{2})$    

$\Rightarrow$   $|\alpha|^{2}-1=-2+8|\alpha|^{2}$

$\Rightarrow$       $7|\alpha|^{2}=1$

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