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 $|\alpha |$ is equal to

• 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 $\frac{1}{\sqrt{7}}$

The intersection of circles, the basic concept is to use equations simultaneously and properties of modulus 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.
$\therefore |\alpha -z_0{|}^2=r^{2}and{|\frac{1}{\alpha }-z_0|}^2=4r^2\Rightarrow (\alpha -z_0)(\overline{\alpha }-\overline{z_0})=r^2\Rightarrow |\alpha {|}^2-z_0\overline{\alpha }-\overline{z_0}\alpha +|z_0{|}^2=r^2$ ...(1)
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, $|\alpha {|}^2=\alpha .\overline{\alpha }\Rightarrow \frac{1}{|\alpha {|}^2}-\frac{z_0}{|\alpha {|}^2}\overline{\alpha }-\frac{\overline{z_0}\alpha }{|\alpha {|}^2}+|z_0{|}^2=4r^2\Rightarrow 1-z_0\overline{\alpha }-\overline{z_0}\alpha +|\alpha {|}^2|z_0{|}^2=4r^2|\alpha {|}^2$ ...(2)
On subtracting Eqs. (1) and (2), we get
$(|\alpha {|}^2-1)+|z_0{|}^2(1-|\alpha {|}^2)=r^2(1-4|\alpha {|}^2)\Rightarrow (|\alpha {|}^2-1)(1-|z_0{|}^2)=r^2(1-4|\alpha {|}^2)\Rightarrow (|\alpha {|}^2-1)(1-\frac{r^2+2}{2})=r^2(1-4|\alpha {|}^2)Given,|z_0{|}^2=\frac{r^2+2}{2}(|\alpha {|}^2-1).\left(\frac{-r^2}{2}\right)=r^2(1-4|\alpha {|}^2)(|\alpha {|}^2-1)=-2+8|\alpha {|}^{2}7|\alpha {|}^2=1\therefore |\alpha |=\frac{1}{\sqrt{7}}$