Suppose that instead of parametrizing the circle by arc length $\theta$, so that $(\cos\theta,\sin\theta)$ is a typical point on the circle, one parametrizes it thus:
$$
t\mapsto \left(\frac{1-t^2}{1+t^2}, \frac{2t}{1+t^2}\right)\text{ for }t\in\mathbb{R}\cup\{\infty\}. \tag{1}
$$
The parameter space is the one-point compactification of the circle, i.e. there's just one $\infty$, which is at both ends of the line $\mathbb{R}$, rather than two, called $\pm\infty$. So $\mathbb{R}\cup\{\infty\}$ is itself topologically a circle, and $\infty$ is mapped to $(-1,0)$.
Now do some geometry: let $t$ be the $y$-coordinate of $(0,t)$, and draw a straight line through $(-1,0)$ and $(0,t)$, and look at the point where that line intersects the circle. That point of intersection is just the point to which $t$ is mapped in $(1)$.
Later edit: an error appears below. I just noticed I did something dumb: the mapping between the circle and the line $y=1-x$ that associates a point on that line with a point on that circle if the line through them goes through $(0,0)$ is not equivalent to the one in $(1)$ because the center of projection is the center of the circle rather than a point on the circle.
end of later edit
This mapping is in a sense equivalent to the one you propose: I think you can find an affine mapping from $t$ to your $x$ on the line $y=-x+1$, such that the point on the circle to which $t$ is mapped and the point on the circle to which $x$ is mapped are related by linear-fractional transformations of the $x$- and $y$-coordinates.
The substitution
$$
\begin{align}
(\cos\theta,\sin\theta) & = \left(\frac{1-t^2}{1+t^2}, \frac{2t}{1+t^2}\right) \\[10pt]
d\theta & = \frac{2\,dt}{1+t^2}
\end{align}
$$
is the Weierstrass substitution, which transforms integrals of rational functions of sine and cosine, to integrals of simply rational functions. I'm pretty sure proposed mapping from the $(x,y=-x+1)$ to the circle would accomplish the same thing.
