Does the circle group given by drawing a parallel chord through the identity match the complex multiplicative unit circle group? I've been learning about homeomorphisms on the circle and was watching this excellent algebraic topology video covering some basics of projective geometry, and describing a group structure on the circle which is underpinned by Pascal's theorem.
The group product $X\star Y=Z$ is arrived at by drawing a chord through $X$ and $Y$ and then drawing a parallel chord through the identity, and where that chord meets the circle is the product $Z$.  It's not immediately obvious (at least to me) that drawing a unit circle in the complex plane and setting the group identity to be $1+0i$ makes $(S^1,\star)$ the standard multiplicative unit circle group in the complex plane $(S^1,\times)$.  But picking sample products such as $i\star i=-1$ and in fact any of the points of the compass, does yield matching results for $\star$ and $\times$.
I was able to satisfy myself that results matched for a good selection of elements of the Prufer 2-group, which being dense in the circle and would appear to order-embed, would determine this must be the same group.  Is it the exact same group and product?
 A: Let $z_j = x_j + i y_j  \in S^1$ ($j = 1,2$) be two elements on the unit circle.
In order to show that the two multiplications are the same it suffices to show that the slope of the chord between $z_1$ and $z_2$ coincides with the slope of the chord between $z_1 z_2 = (x_1 x_2 - y_1 y_2) + i (x_1 y_2 + x_2 y_1)$ and $1 + 0i$.
That means we have to show that
$$ (x_1 - x_2) (x_1 y_2 + x_2 y_1) = (y_1 - y_2) (x_1 x_2 - y_1 y_2 - 1)$$
and with the help of $x_j^2 + y_j^2 = 1$ for $j = 1,2$, this is easily done.
A: I know this already has an answer, but there is a more geometric way of thinking about it that you might find nice.

*

*The points $1, x, y, z$ form a cyclic quadrilateral. To avoid confusing notation, call these points $A, B, C, D$ respectively. Assume $x$ comes before $y$ as you traverse the circle anticlockwise.

*Since the line $\vec{AD}$ is parallel to $\vec{BC}$, this quadrilateral is a trapezium.

*Since it's a cyclic quadrilateral, $A+C = B+D = \pi$. Since it's a trapezium, $A+B = C+D = \pi$.

*It follows the whole diagram is symmetric under reflection in the perpendicular bisector of AD (which is also the perpendicular bisector of BC).

*If you add O, the center of the circle, to the diagram, then the desired fact to prove is

$$\arg x + \arg y = \arg z$$
i.e.
$$\angle AOB + \angle AOC = \angle AOD$$
But it's obvious (here we use the anticlockwise orientation of $ABCD$) that $\angle AOD = \angle AOC + \angle COD$, and by the symmetry $\angle COD = \angle AOB$. Hence,
$$\angle AOD = \angle AOC + \angle AOB$$
which is what we wanted to prove.
A: Too long for a comment.
Consider the reducible degree three curve $${\cal C}: z(x^2+y^2-z^2)=0$$ in ${\Bbb P}^2_{\Bbb R}$ with $O$ the point $(x:y:z)=(1:0:1).$
Let $A,B,C,$ and $O=(1,0)$ be points on $x^2+y^2-1=0$ in ${\Bbb R}^2={\Bbb A}^2_{z=1}\subset{\Bbb P}^2.$
$A$ and $B$ in ${\Bbb P}^2$ define a line that intersects ${\cal C}$ in a point I call $\infty_{AB}.$ Then proceed as with the elliptic curve group law:
$$A+B+\infty_{AB} =0$$ and similarly $$\infty_{OC} + O +C=0$$ remembering from projective geometry that $AB$ parallel to $OC$ means that $\infty_{AB}=\infty_{OC},$ we have $$A+B=-\infty_{AB}=-\infty_{OC}=O+C.$$ If we let $O$ be the identity for $+,$ we get $$A+B=C.$$
