Let the complex numbers \(\alpha\) and \(\left(\frac{1}{\bar{\alpha}}\right)\) lie on circles…
Let the complex numbers \(\alpha\) and \(\left(\frac{1}{\bar{\alpha}}\right)\) 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 \(z_0=x_0+i y_0\) satisfies the equation \(2\left|z_0\right|^2=\) \(r^2+2\), then \(|\alpha|=\)
\(\frac{1}{\sqrt{2}}\)
\(\frac{1}{2}\)
\(\frac{1}{\sqrt{7}}\)
\(\frac{1}{3}\)
Solution
As point \(\alpha\) lies on the circle
\(\begin{aligned}
& \left(x-x_0\right)^2+\left(y-y_0\right)^2 =r^2 \\
\therefore & \left|\alpha-z_0\right|^2=r^2, \text { where } z_0 =x_0+i y_0 \\
\Rightarrow & |\alpha|^2+\left|z_0\right|^2-\left(\alpha \bar{z}_0+\bar{\alpha} z_0\right) =r^2 \quad \ldots (i)
\end{aligned}\)
\(\because \frac{1}{\bar{\alpha}}\) lies on the circle \(\left(x-x_0\right)^2+\left(y-y_0\right)^2=4 r^2\)
\(\therefore \quad\left|\frac{1}{\bar{\alpha}}-z_0\right|^2=4 r^2\)
\(\begin{aligned}
& \Rightarrow \quad \frac{1}{|\alpha|^2}+\left|z_0\right|^2-\left(\frac{\alpha \bar{z}_0}{|\alpha|^2}+\frac{\bar{\alpha} z_0}{|\alpha|^2}\right)=4 r^2 \\
& \Rightarrow \quad 1+\left|z_0\right|^2|\alpha|^2-\left(\alpha \bar{z}_0+\bar{\alpha} z_0\right)=4 r^2|\alpha|^2 \quad \ldots (ii)
\end{aligned}\)
By subtracting Eqs. (i) and (ii), we get
\(\begin{array}{cc}
& 1-|\alpha|^2-\left|z_0\right|^2\left(1-|\alpha|^2\right)=r^2\left(4|\alpha|^2-1\right) \\
\Rightarrow & \left(|\alpha|^2-1\right)\left(\left|z_0\right|^2-1\right)=r^2\left(4|\alpha|^2-1\right) \\
\because & \left|z_0\right|^2=\frac{r^2+2}{2}, \text { we get }
\end{array}\)
\(\begin{aligned}
& \left(|\alpha|^2-1\right) \frac{r^2}{2} =r^2\left(4|\alpha|^2-1\right) \\
\Rightarrow & |\alpha|^2-1 =8|\alpha|^2-2 \\
\Rightarrow & 7|\alpha|^2=1 \Rightarrow|\alpha| =\frac{1}{\sqrt{7}}
\end{aligned}\)