Question

Let complex numbers $\alpha \text{ and }{\displaystyle \frac{1}{\alpha }}$ lie on circles ${\left(x-x_o\right)}^2+{\left(y-y_o\right)}^2=r^2$ and ${\left(x-x_o\right)}^2+{\left(y-y_o\right)}^2=4r^2$, respectively. If $z_o=x_o+i{y}_o$ satisfies the equation, $2{\left|z_o\right|}^2=r^2+2$, then $\left|\alpha \right|$ =
$\left(a\right){\displaystyle \frac{1}{\sqrt{2}}}$
$\left(b\right){\displaystyle \frac{1}{2}}$
$\left(c\right){\displaystyle \frac{1}{\sqrt{7}}}$
$\left(d\right){\displaystyle \frac{1}{3}}$

Answer 1

Hint: In this particular question use the concept that in the standard equation of the circle ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$, (h, k) and r are the center and the radius of the circle respectively and in complex plane we can write the equation of the circle, $\left|z-z_o\right|=r$ where z and $z_o$ both are the complex numbers and r is the radius of the circle, so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given data:
Complex numbers $\alpha \text{ and }{\displaystyle \frac{1}{\alpha }}$ lie on circles ${\left(x-x_o\right)}^2+{\left(y-y_o\right)}^2=r^2$ and ${\left(x-x_o\right)}^2+{\left(y-y_o\right)}^2=4r^2$.
So the equations of the circles in the complex plane is written as,
$\Rightarrow \left|z-z_o\right|=r$ and $\left|z-z_o\right|=2r$
Now $\alpha \text{ and }{\displaystyle \frac{1}{\alpha }}$ which represents the complex number and lie on the circle so it satisfies the above equation respectively, so we have,
$\Rightarrow \left|\alpha -z_o\right|=r$, and $\left|{\displaystyle \frac{1}{\alpha }}-z_o\right|=2r$
$\Rightarrow \left|\alpha -z_o\right|=r$, and ${\displaystyle \frac{\left|1-\alpha z_o\right|}{\left|\alpha \right|}}=2r$
$\Rightarrow \left|\alpha -z_o\right|=r$, and $\left|1-\alpha z_o\right|=2r\left|\alpha \right|$
Now squaring on both sides in both of the above equation we have,
$\Rightarrow {\left|\alpha -z_o\right|}^2=r^2$, and ${\left|1-\alpha z_o\right|}^2=4r^2{\left|\alpha \right|}^2$
Now as we know that ${\left|a-b\right|}^2=\left(a-b\right)\left(\overline{a-b}\right)=\left(a-b\right)\left(\overline{a}-\overline{b}\right)$ and ${\left|1-ab\right|}^2=\left(1-\overline{a}b\right)\left(1-a\overline{b}\right)$ where a and b both represents the complex numbers so use these properties in the above equation we have,
$\Rightarrow \left(\alpha -z_o\right)\left(\overline{\alpha -z_o}\right)=r^2$ , and $\left(1-\overline{\alpha }{z}_o\right)\left(1-\alpha {\overline{z}}_o\right)=4r^2{\left|\alpha \right|}^2$
$$\Rightarrow \left(\alpha -z_o\right)\left(\overline{\alpha }-{\overline{z}}_o\right)=r^2$$ , and $\left(1-\overline{\alpha }{z}_o\right)\left(1-\alpha {\overline{z}}_o\right)=4r^2{\left|\alpha \right|}^2$
Now simplify it we have,
$$\Rightarrow \left(\alpha -z_o\right)\left(\overline{\alpha }-{\overline{z}}_o\right)=r^2$$
$$\Rightarrow \alpha \overline{\alpha }-\alpha {\overline{z}}_o-z_o\overline{\alpha }+z_o{\overline{z}}_o=r^2$$
$$\Rightarrow \alpha {\overline{z}}_o+z_o\overline{\alpha }=\alpha \overline{\alpha }+z_o{\overline{z}}_o-r^2$$
$$\Rightarrow \alpha {\overline{z}}_o+z_o\overline{\alpha }={\left|\alpha \right|}^2+{\left|z_o\right|}^2-r^2$$..................... (1), $$\left[\because \alpha \overline{\alpha }={\left|\alpha \right|}^2,z_o{\overline{z}}_o={\left|z_o\right|}^2\right]$$
And
$\Rightarrow \left(1-\overline{\alpha }{z}_o\right)\left(1-\alpha {\overline{z}}_o\right)=4r^2{\left|\alpha \right|}^2$
$\Rightarrow 1-\alpha {\overline{z}}_o-z_o\overline{\alpha }+\alpha \overline{\alpha }{z}_o{\overline{z}}_o=4r^2{\left|\alpha \right|}^2$
$\Rightarrow 1-\alpha {\overline{z}}_o-z_o\overline{\alpha }+{\left|\alpha \right|}^2{\left|z_o\right|}^2=4r^2{\left|\alpha \right|}^2$
$\Rightarrow \alpha {\overline{z}}_o+z_o\overline{\alpha }=1+{\left|\alpha \right|}^2{\left|z_o\right|}^2-4r^2{\left|\alpha \right|}^2$................... (2)
Now as we see that the LHS of the equation (1) and (2) are same so equate RHS of the equation (1) and (2) we have,
$$\Rightarrow {\left|\alpha \right|}^2+{\left|z_o\right|}^2-r^2=1+{\left|\alpha \right|}^2{\left|z_o\right|}^2-4r^2{\left|\alpha \right|}^2$$.............. (3)
Now it is given that $2{\left|z_o\right|}^2=r^2+2$, $\Rightarrow {\left|z_o\right|}^2={\displaystyle \frac{r^2+2}{2}}$ so substitute this value in equation (3) we have,
$$\Rightarrow {\left|\alpha \right|}^2+{\displaystyle \frac{r^2+2}{2}}-r^2=1+{\left|\alpha \right|}^2\left({\displaystyle \frac{r^2+2}{2}}\right)-4r^2{\left|\alpha \right|}^2$$
Now simplify it we have,
$$\Rightarrow {\left|\alpha \right|}^2-{\displaystyle \frac{r^2}{2}}+1=1+{\displaystyle \frac{{\left|\alpha \right|}^2{r}^2}{2}}+{\left|\alpha \right|}^2-4r^2{\left|\alpha \right|}^2$$
Now cancel out the same terms with same sign from LHS and RHS we have,
$$\Rightarrow -{\displaystyle \frac{r^2}{2}}={\displaystyle \frac{{\left|\alpha \right|}^2{r}^2}{2}}-4r^2{\left|\alpha \right|}^2$$
Now divide by $r^2$ throughout we have,
$$\Rightarrow -{\displaystyle \frac{1}{2}}={\displaystyle \frac{{\left|\alpha \right|}^2}{2}}-4{\left|\alpha \right|}^2$$
$$\Rightarrow -{\displaystyle \frac{1}{2}}=-{\displaystyle \frac{7{\left|\alpha \right|}^2}{2}}$$
$$\Rightarrow 7{\left|\alpha \right|}^2=1$$
$$\Rightarrow {\left|\alpha \right|}^2={\displaystyle \frac{1}{7}}$$
Now take square root on both sides we have,
$$\Rightarrow \left|\alpha \right|=\sqrt{{\displaystyle \frac{1}{7}}}={\displaystyle \frac{1}{\sqrt{7}}}$$
So this is the required answer.
Hence option (c) is the correct answer.


Note: Whenever we face such types of questions the key concept we have to remember is that always recall that ${\left|a-b\right|}^2=\left(a-b\right)\left(\overline{a-b}\right)=\left(a-b\right)\left(\overline{a}-\overline{b}\right)$ and ${\left|1-ab\right|}^2=\left(1-\overline{a}b\right)\left(1-a\overline{b}\right)$ where a and b both represents the complex numbers, so according to this property simplify the equation as above and then equate the RHS of the above equations as LHS are same then again simplify the equation using the given condition in the question as above we will get the required answer.