Let $w=\varepsilon\beta(t)-i\sqrt{\beta(t)^2-1}$, where $\beta(t)=\cosh t$ and $\varepsilon >0$.
the parametric function is defined as $x+iy=\frac{2w}{|w|^2+1}$ and $z=\frac{|w|^2-1}{|w|^2+1}$.
Question is when $t\in(-\infty,\infty)$, does the curve $(x(t),y(t),z(t))$ defines a circle?
It is easy to show that $(x(t),y(t),z(t))$ is on a unit sphere, so we need to prove they are on a plane.
I don't think it is on circle, for $t\in Real$, $$\omega=\varepsilon \cosh t - i \sinh t\\x=\frac{2\varepsilon \cosh t}{(\varepsilon^2+1)\cosh ^2 t},y=\frac{-2\sinh t}{(\varepsilon^2+1)\cosh ^2 t},z=1-\frac{2}{(\varepsilon^2+1)\cosh ^2 t}$$ so maybe it is $$x+i y=\frac{2\omega}{\sqrt{|\omega|^2+1}},z=\sqrt{\frac{|\omega|^2-1}{|\omega|^2+1}}$$ then $$x=\frac{2\varepsilon}{\sqrt{(\varepsilon^2+1)}},y=\frac{-2\sinh t}{\sqrt{(\varepsilon^2+1)}\cosh t},z=\sqrt{1-\frac{2}{(\varepsilon^2+1)\cosh ^2 t}}\\x^2+y^2+z^2=3$$ on a sphere. Then use $t=0$, and $t\to \infty$, the points $(\frac{2 \varepsilon}{\sqrt{(\varepsilon ^2 + 1)}},0,\sqrt{\frac{\varepsilon ^2 - 1}{\varepsilon ^2 + 1}})$ and $(\frac{2 \varepsilon}{\sqrt{(\varepsilon ^2 + 1)}}, \frac{-2}{\sqrt{\varepsilon ^2+1}},1)$ are on the curve, use it with the center, one can find the normal of the plane(if it is a plane), and then check the $P(t)\cdot N=0$, However, if $x$ is not moving, they must be on a $y-z$ plane.