How is $x(t)=r*\frac{1+it}{1-it}$ with $-\infty < t < \infty$ a parametrisation of the circle with centre zero and radius $r$? (Remmert, Schumacher)

Question

In Remmert, Schumacher: Theory of functions (German Edition 6.2.3 p.156) it is written that

$x(t)=r*\frac{1+it}{1-it}$ with $-\infty < t < \infty$ is a parametrisation of the circle with centre zero and radius $r$, as it is the second point of intersection of a straight line that goes through $(-r,0)$ and the boundary of the circle.

I assume that the circle can be written either in form of the parametrisation $y(t)=r*exp(it)$ or the function $y(t)=\sqrt(r^2-t^2)$ and then I should be able to get $x(t)=r*\frac{1+it}{1-it}$ as intersection, but unfortunately I am not.

I appreciate any help!

Answer 1

Better write $z(t)=r \frac{1+it}{1-it}$ then $$z \bar z= r^2 =|z|^2 \implies |z|=r \implies x^2+y^2=r^2$$ Because $z=x+iy$ and $\bar z=x-iy$.