What is the meaning of inner product of two circles? I was reading this post about orthogonal circles (here) and in the question body something about inner product of circles was used.. it seemed quite mysterious but still interesting. I know of inner products of vector which is simply the dot product but what exactly is the inner product of circles? Does it having an associated geometric meaning?
 A: I checked Iversen's book "Hyperbolic Geometry". This formula is indeed the Lorentzian inner product restricted to points of the 1-sheeted hyperboloid $M$. Here is how the map from circles to points in the  1-sheeted hyperboloid works. Consider a circle $C$ in the Euclidean plane. If you think of the upper half-space model $U$ of the hyperbolic 3-space ${\mathbb H}^3$:
$$
U=\{(x_1, x_2, x_3): x_3> 0\}
$$
then $C$ is the intersection of the closure of a unique hyperbolic plane $H_C\subset U$ with the Euclidean plane $\{x_3=0\}$. Now, you switch to the Lorentzian model of the hyperbolic 3-space. Then ${\mathbb H}^3$  becomes the upper sheet $S$ of the 2-sheeted hyperboloid
$$
\{v: \langle v, v\rangle=-1\},
$$
where $\langle v, w\rangle$ denotes the Lorentzian inner product in the Lorentzian space $R^{3,1}$. Then hyperbolic planes in ${\mathbb H}^3$ correspond to the intersection of 3-dimensional linear subspaces $V\subset R^{3,1}$ with $S$. The assumption that the intersection $V\cap S$ is nonempty translates into the condition that the Lorentzian perpendicular vector $u$ to $V$ is space-like, i.e.
$$
\langle u, u\rangle > 0
$$
WLOG, we can assume then that $u$ is a Lorentzian unit vector.
I will consider the circle $C$ oriented, then $H_C$ is oriented as well, hence $V$ is also oriented. This determines the direction of $u$, hence, determines $u$ uniquely. The vector $u$ then belongs to the 1-sheeted hyperboloid $M$ in $R^{3,1}$.
Thus, we obtain the map $C\mapsto u=u_C$, sending oriented Euclidean circles to points of the 1-sheeted hyperboloid $M$. (This map is 1-1 but not onto: The mossing points correspond to Euclidean lines which can be regarded as circles of infinite radius.) Next, each circle $C$ in the plane is given by the quadratic equation
$$
b (x\cdot x) - 2(x\cdot f) + a=0,  
$$
where $\cdot$ is the Euclidean dot product, $f\in {\mathbb R}^2$ and $a, b\in {\mathbb R}$. This equation is not quite unique since we can multiply $a, b, f$ by the same scalar without changing the zero set of the quadratic polynomial. So, one normalizes
$$
|f|^2 -ab=1.
$$
Then the "inner product" of two circles becomes
$$
\langle C_1, C_2\rangle = (f_1\cdot f_2) - \frac{1}{2}( a_1 b_2 + b_1 a_2). 
$$
This is the formula from the linked MSE question.
Now, one has to do a computation which yields:
$$
\langle u_{C_1}, u_{C_2}\rangle= \langle C_1, C_2\rangle, 
$$
where the inner product on the left is Lorentzian.
Another geometric interpretation of this inner product is the negative of the cosine of the angle between the circles.
Needless to say, all this also works in higher dimensional Euclidean and hyperbolic spaces, just one has to replace circles by spheres.
A: We may stereographically project $\mathbb{R}^n$ to $S^n\subseteq\mathbb{R}^{1+n}$ given by $u\mapsto\big(\frac{1-|u|^2}{1+|u|^2},\frac{2u}{1+|u|^2}\big)$.
Let's say you're already slightly familiar with pseudo-Euclidean spaces $\mathbb{R}^{p,q}$, and know that the null cone consists of rays through a copy of $S^{p-1}\times S^{q-1}$; then you also know $S^n$ may be identified with $\{1\}\times S^n$ in the null cone and this forms a complete set of representatives for the projectivized null cone (i.e. null vectors mod rescaling). Then we may rescale by to write $\mathbb{R}^n\to S^n\to \mathbb{R}^{1,1+n}$ as
$$ u \mapsto \left(\frac{1-|u|^2}{1+|u|^2},\frac{2u}{1+|u|^2}\right) \mapsto \left(\frac{1+|u|^2}{2},\frac{1-|u|^2}{2},u\right). $$
Indeed, this map $\phi$ can be considered a form of stereographic projection. Now let $n=2$.
Note the pseudo- inner product $\xi\cdot \phi(u)$ is quadratic in $u$, and similarly the equation of a circle in the plane centered at $p$ of radius $r$, $|u-p|^2=r^2$, is similarly quadratic in $u$. Thus, for any choice of circle $C(p,r)$, there should be a corresponding $\xi$ for which $|u-p|^2=r^2 ~\Leftrightarrow~ \xi\cdot\phi(u)=0$. Explicitly, solving
$$ |u|^2-2p\cdot u + (|p|^2-r^2) = 0 $$
$$ \iff $$
$$ -\xi_1\frac{1+|u|^2}{2} +\xi_2\frac{1-|u|^2}{2}+\zeta\cdot u =0 $$
yields (after rescaling)
$$ \xi(p,r)=\left(\frac{1+|p|^2-r^2}{2},\frac{1-|p|^2+r^2}{2},p\right). $$
Of course, you're curious what the inner product of two circles is, so you compute
$$ \xi(p_1,r_1)\cdot\xi(p_2,r_2)=\frac{1}{2}(r_1^2+r_2^2-|p_1-p_2|^2). $$
If the two circles intersect, then the terms of this expression are the same as those that appear in the law of cosines for the triangle vertexed at the circle centers and one of their points of intersection. Indeed, the angle $\varphi$ between the legs of this triangle is the (acute) angle between the circles' tangent lines at their intersection, because the triangle legs are just the tangent lines rotated by a right angle. Therefore,
$$ \xi(p_1,r_1)\cdot\xi(p_2,r_2)=r_1r_2\cos\varphi $$
by the law of cosines.
It is also noteworthy that $\xi(p,r)\cdot\xi(p,r)=r^2$, so the above is reminiscent of the  norm-angle formula for the usual Euclidean inner product. This observation can also help generalize $\xi$ to a projection $\eta:\mathbb{R}^n\to\mathbb{R}^{2,1+n}$.
You may also look up "power of a point" or "Darboux product." The projection $\xi$ is the topic of "Mobius geometry" which leads to $\eta$ in "Lie sphere geometry." One source for this is Cecil.
