Is this a group on the circle? Is there a way the set of real points on any circle be made into a group?
I was trying to form a group from the real points of a circle as below:


*

*The identity element will be the point $(0,1)$. 

*To find the product of any two points $A$ and $B$, draw a straight line through them (so we have the line $\overline{AB}$), then draw a line parallel to $\overline{AB}$ that passes through $(0,1)$, the identity point. The point where it intersects the circle is the product of $A$ and $B$.

*The inverse of any point is the point directly opposite that point horizontally. Example the inverse of the point $(1,0)$ is the point $(-1,0)$. With this setting, we see the inverse of the identity $(0,1)$ is itself. Also, the inverse of the $(0,-1)$ is itself.
 A: This is indeed a group! 
Moreover, this same group structure can be put onto any conic -- not just a circle. The reason why this is a group and why this works with conics is due to where the proof for associativity comes from: Pascal's theorem. As you have already discovered, the rest of the group axioms come more-or-less for free from the definition of the operation. 
You may be interested in two papers on the subject:


*

*F. Lemmermeyers' Conics -- A Poor Man's Elliptic Curves

*S. Shirali's Groups Associated with Conics
As you will see in Lemmermeyers' paper, you can endow elliptic curves with a similar structure. You can read more about the elliptic curve group here or here. The group is also mentioned in some algebraic geometry texts as well.
Edit: N. J. Wildberger mentions this group in a lecture of his on Algebraic Topology and talks some about the proof of associativity. 
A: This will give a group and it will be the usual circle group. This is the group on the set $\{e^{i\phi}\mid \phi\in[0,2\pi)\}$ with complex multiplication as group operation. Alternatively it can be written as $\Bbb R/2\pi\Bbb Z$ as states in Evargalo's answer.
Proof.
Lets say the two points of the circle are given as $e^{i\phi}$ and $e^{i\psi}$. The corresponding line through the $1$ according to your description is given by
$$g(t)=1+t(e^{i\phi}-e^{i\psi}).$$
We are looking for the (if possible) non-zero value of $t$ for which $\|g(t)\|=1$. We can agree that there is at most one from the geometric intuition on this problem. So let me show you that
$$t^*=\frac{e^{i(\phi+\psi)}-1}{e^{i\phi}-e^{i\psi}}$$
is this value. This will give you $g(t^*)=e^{i(\phi+\psi)}=e^{i\phi}e^{i\psi}$. All we need to show is that $t^*$ is real. For this, multiply nominator and denominator with $e^{-1/2i\phi}e^{-1/2i\psi}$ to obtain
$$t^*=\frac
{e^{1/2i\phi}e^{1/2i\psi}-e^{-1/2i\phi}e^{-1/2i\psi}}
{e^{1/2i\phi}e^{-1/2i\psi}-e^{-1/2i\phi}e^{1/2i\psi}}.$$
Setting $z=e^{1/2i\phi}e^{1/2i\psi}$ and $w=e^{1/2i\phi}e^{-1/2i\psi}$ with their complex conjugates $\bar z$ and $\bar w$, we see that
$$t^*=\frac{z-\bar z}{w-\bar w}
=\frac{2i\cdot\mathrm{Im}(z)}{2i\cdot\mathrm{Im}(w)}
=\frac{\mathrm{Im}(z)}{\mathrm{Im}(w)}$$
which is real. $\;\square$
To be completely rigorous you will have to argue seperately for the case $t^*=\mathrm{Im}(z)=0$ and $\mathrm{Im}(w)=0$, but I hope this will do it for now.

I wrote a question motivated on this one, and you can see from the answer that no matter what you try, as soon as you develop a (topological) group structure on $S^1$, it will be the usual circle group structure.
A: If you use the argument-modulus representation of the points of the circle (choosing the center of the circle as the origin and its radius as the unit), then:
$\mathcal{C}=\{e^{i\theta}, \theta\in[0,2\pi)\}$ is isomorphic to the additive group $\mathbb{R}/2\pi$
