We work in the Extended Complex Plane: $ \mathbb{C} \cup (\infty)$.

Basically, say we have two points, $z_1$ and $z_2$.

It can be shown that, on stereographic projection of the Riemann Sphere onto the Extended Complex , spherical lines are either straight-lines through the origin, (in the instance that our spherical line passes through the North Pole) or circles of the form: $\{z: |z - w|^2 = |w|^2+1\}$, for some $w \in \mathbb{C}$.

Similarly, if we work in the Poincare Disk Model of Hyperbolic geometry, hyperbolic lines in the complex plane are either straight lines through the origin, (if $z_1, 0, z_2$ are collinear) or take the form, $\{z: |z - w|^2 = |w|^2-1\}$, for some $w \in \mathbb{C}$ with $|w| >1$.

The task, is essentially, to find $w$ in terms of $z_1$ and $z_2$.

Now, it should be fairly straightforward, as far as solving a pair of simultaneous equations go, in order to pin-point $w$. However, I find, after some playing around that it quickly devolves into a recalcitrant algebraic mess and I lose much intuition.

For instance, in the hyperbolic case, if I let $z_1 = x_1 + iy_1$ and $z_2 = x_2 + iy_2$, I get

$w = a( 1- i\left(\frac{x_2 -x_1}{y_2-y_1}\right)) + \frac{i}{2}\left(\frac{|z_2|^2 - |z_1|^2}{y_2 - y_1} \right)$, where,

$a = \frac{|z_1|^2y_2 + |z_1|^2y_1 + y_2 - y_1}{2(x_1y_2 - x_2y_1)}$, which just seems painful to simplify.

And so, my question is, am I simply to persist through all this painful algebra, or is there something I'm just not seeing? And is there any more intuitive way to look at where these quantities come from? An obvious answer is to consider the inverse image on the Riemann Sphere and see what happens, but, ah well.

In any instance, thank you!


Deriving additional points

There is something you have not mentioned, and which might make things easier. The way I understand your question, you are essentially dealing with the situation where you have two points and want to find the center of the Euclidean circle which corresponds to the geodesic through these two points.

In both cases, you can compute two more points on the same circle. In spherical geometry you can project one of the points onto the sphere, take its antipode and project that back to the plane. In the Poincaré model of hyperbolic geometry, you can invert the points in the unit circle. So now you have four points, any three of which define the circle you're after. You can then continue to construct the center using perpendicular bisectors and so on.

Hyperbolic case

I'll start with the hyperbolic case, since I consider that easier. If you have $(x_1,y_1)$ as one of the points, its inverse in the unit circle is $\frac1{x_1^2+y_1^2}(x_1,y_1)$ so their midpoint is $\frac{x_1^2+y_1^2+1}{2(x_1^2+y_1^2)}(x_1,y_1)$ and the perpendicular bisector is $2x_1\cdot x + 2y_1\cdot y = x_1^2+y_1^2+1$. Changing the index you obtain a similar perpendicular bisector for $z_2$. These two lines intersect in the point $w$ for the hyperbolic case, which can be written as a complex number like this:

$$w=\frac{ \begin{vmatrix}x_1^2+y_1^2+1 & y_1 \\ x_2^2+y_2^2+1 & y_2\end{vmatrix} +i\begin{vmatrix}x_1 & x_1^2+y_1^2+1 \\ x_2 & x_2^2+y_2^2+1\end{vmatrix} }{2\begin{vmatrix}x_1&y_1\\x_2&y_2\end{vmatrix}}=\frac{ \begin{vmatrix}\lvert z_1\rvert^2+1 & y_1 \\ \lvert z_2\rvert^2+1 & y_2\end{vmatrix} +i\begin{vmatrix}x_1 & \lvert z_1\rvert^2+1 \\ x_2 & \lvert z_2\rvert^2+1\end{vmatrix} }{2\begin{vmatrix}x_1&y_1\\x_2&y_2\end{vmatrix}}$$

This simple application of Cramer's rule is probably as much structure as you can expect from this expression. It likely won't become much easier if you try to simplify it further. In particular, you can't cancel any terms, since $w$ has to become $\infty$ if $z_1$ and $z_2$ are collinear with $0$, which is what the denominator signifies. The expression is also very similar to what you posted in your question, so you were pretty close already, and perhaps you don't consider all of this any simplification at all.

Spherical case

Connecting $(x_1,y_1,0)$ with $(0,0,1)$ and intersecting that with the unit sphere, you obtain $\frac{1}{x_1^2+y_1^2+1}(2x_1,2y_1,x_1^2+y_1^2-1)$. The antipode of this is its negative, and projecting that back to the plane you end up at $\frac{-1}{x_1^2+y_1^2}(x_1,y_1)$. So the antipode on the sphere is the negative of the inverse in the unit circle. Perhaps not at all surprising, and probably you already knew this. The midpoint now is $\frac{x_1^2+y_1^2-1}{2(x_1^2+y_1^2)}(x_1,y_1)$ and the perpendicular bisector is $2x_1\cdot x+2y_1\cdot y=x_1^2+y_1^2-1$ which means that the center will be

$$w=\frac{ \begin{vmatrix}x_1^2+y_1^2-1 & y_1 \\ x_2^2+y_2^2-1 & y_2\end{vmatrix} +i\begin{vmatrix}x_1 & x_1^2+y_1^2-1 \\ x_2 & x_2^2+y_2^2-1\end{vmatrix} }{2\begin{vmatrix}x_1&y_1\\x_2&y_2\end{vmatrix}}=\frac{ \begin{vmatrix}\lvert z_1\rvert^2-1 & y_1 \\ \lvert z_2\rvert^2-1 & y_2\end{vmatrix} +i\begin{vmatrix}x_1 & \lvert z_1\rvert^2-1 \\ x_2 & \lvert z_2\rvert^2-1\end{vmatrix} }{2\begin{vmatrix}x_1&y_1\\x_2&y_2\end{vmatrix}}$$

  • $\begingroup$ Yes, I realised. I ended up bashing through the algebra stubbornly enough, regardless, and got more or less, exactly what you did. In any instance, thank you! $\endgroup$ – AlpArslan Apr 16 '15 at 20:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.