Suppose we have a circle centered around $(0,1)$ with radius $1$ i.e the circle described by $x^2+(y-1)^2=1$. Now we construct a point on this circle corresponding to a real number by drawing a straight line ($y=kx+2$) from the point $(0,2)$ on the circle to the $x$ axis in the real plane and say that the point where the line intersects the circle is the new point for the point at which the line intersects the real line.
Example : suppose we have the point $(2,0)$ then the straight line passing trough the top of the circle $(0,2)$ and $(2,0)$ is the line $y=-x+2$ which intersects the circle a second time at the point $(1,1)$. Thus we assign the point $(2,0)$ (or simply the number $2$) to be $(1,1)$. (See below an image of two such lines)
More generally if we wish to find the intersection for a number $t \in \mathbb{R} \setminus \{0\}$ we find ( by plugging the formula for a straight line into the circle) an intersection point at $$x=\frac{4t}{4+t^2} \quad y=\frac{-2}{t}\left(\frac{4t}{4+t^2}\right)+2$$
Now my reason for this is that I wish to define a metric d as the (positive) arc-length between the points projected onto the circle. I have checked that one formula for finding the arc length is $"arc-length"= (angle) \times radius$
So using the cosine-formula $$a^2=b^2+c^2-2bc\cos(A)$$ with the angle A centered at the middle of the circle and a, b and c corresponds to the sides of the triangle with vertices at the middle, and at two points of the circle. (see picture below for an explanation) So , to obtain the angle $A$ between two points, we take two points $s$ and $t$ on the real line and then using the distance formula we take their corresponding point on the circle as obtained above to obtain $a$ $$\begin{align*}a & =\sqrt{\left(\frac{4s}{4+s^2}-\frac{4t}{4+t^2}\right)^2+\left(\frac{-2}{s}\left(\frac{4s}{4+s^2}\right)+2-\left(\frac{-2}{t}\left(\frac{4t}{4+t^2}\right)+2\right)\right)^2}\\ &= \sqrt{\left(\frac{4s}{4+s^2}-\frac{4t}{4+t^2}\right)^2+\left(\frac{-8}{4+s^2}+\frac{8}{4+t^2}\right)^2}\end{align*} $$ then $b=c=1$ since this is the radius
so \begin{align*}A & =\arccos\left(\frac{a^2-b^2-c^2}{-2bc} \right) \\ & = \arccos\left(\frac{\left(\frac{4s}{4+s^2}-\frac{4t}{4+t^2}\right)^2+\left(\frac{-8}{4+s^2}+\frac{8}{4+t^2}\right)^2-2}{-2} \right)\end{align*} Which , since our radius is 1, only needs to be put in absolut value bars to obtain the arc-length metric (i.e from above $|arc-length|=|A\times 1|$). That is $$d(s,t)= \left|\arccos\left(\frac{\left(\frac{4s}{4+s^2}-\frac{4t}{4+t^2}\right)^2+\left(\frac{-8}{4+s^2}+\frac{8}{4+t^2}\right)^2-2}{-2} \right)\right|$$ Too check this is a metric:
$d(s,s)=arccos(1)=0$
$d(s,t)=d(t,s)$ due to the fact that one can factor a $-1$ from the square expression.
So the question is if the triangle inequality holds. But I am getting stuck at what to do with the fact that certain points (the negative part of $\mathbb{R}$) is mapped to the left of the circle and the others on the right. Perhaps it is an obvious triangle inequality due to $|x+y|\leq |x|+|y|$? Any help or tips is greatly appreciated!
PS. This question came up when discussing metrics which are not norms, where only the discrete metric came to mind.
Pictures:
M=middle point and all other letters correspond to the variables above. A=angle
example of lines and their corresponding intersection with the circle. the purple line intersects the x axis at 1.25 and the light blue at 1.8182
