All the points in the $\mathrm{set} \left\{S = \frac{\alpha + i}{\alpha - i}; \alpha \in R\right\} \left(i = \sqrt{-1}\right) lie on a$

$\left\{\mathrm{S}=\frac{\mathrm{\alpha }+\mathrm{i}}{\mathrm{\alpha }-\mathrm{i}}, \mathrm{\alpha }\in \mathrm{R}\right\}$

Let $\mathrm{S}=\mathrm{x}+\mathrm{iy}$

$$\begin{gathered} \therefore \mathrm{x}+\mathrm{iy}=\frac{\mathrm{\alpha }+\mathrm{i}}{\mathrm{\alpha }-\mathrm{i}} \\ \Rightarrow \left|\mathrm{x}+\mathrm{iy}\right|=\frac{\left|\mathrm{\alpha }+\mathrm{i}\right|}{\left|\mathrm{\alpha }-\mathrm{i}\right|} \\ \Rightarrow \sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}=\frac{\sqrt{{\mathrm{\alpha }}^2+1}}{\sqrt{{\mathrm{\alpha }}^2+1}} \\ \therefore {\mathrm{x}}^2+{\mathrm{y}}^2=1 \end{gathered}$$

S lies on a circle with centre (0, 0),  radius = 1