Is there any closed formula for the area of the intersection of two circles in the hyperbolic plane $\mathbb{H}^2$? The two circles have radii $R, R'$ and a distance of $d$ between centers. If possible, the curvature should be a parameter (though I am not sure if the curvature affects the area).


I have a version without any integrals: four acute angles related by four equations, but not an easy system to solve. Then your area is a simple function of those angles. If it will help, i can type it in. By the way, the calculation is intrinsic. The upper half plane or disc just makes everything worse.

Meanwhile, looking at your hyperbolic network question, you should become aware that the hyperbolic plane is the natural setting for any extremely regular network, one where every edge is the same length, each node is connected to the same number of neighbors, spaced at equal angles. See material on Ford Circles, but especially http://en.wikipedia.org/wiki/Uniform_tilings_in_hyperbolic_plane

EDIT, Sunday 23 Feb.

$$ \frac{\sin \alpha}{\sinh a} = \frac{\sin \beta}{\sinh b} =\frac{\sin \gamma}{\sinh c} $$

$$ \cos \alpha = \frac{\cosh b \cosh c - \cosh a}{\sinh b \sinh c } $$

$$ \cosh a = \frac{\cos \alpha + \cos \beta \cos \gamma}{ \sin \beta \sin \gamma} $$

IF $\gamma = \pi / 2,$ $$ \cosh c = \cosh a \cosh b = \cot \alpha \cot \beta $$

$$ \cosh a = \frac{\cos \alpha}{\sin \beta}, \; \; \sin \alpha = \frac{\sinh a}{\sinh c}, $$

$$ \cos \alpha = \frac{\tanh b}{\tanh c}, \; \; \tan \alpha = \frac{\tanh a}{\sinh b} $$

IF also two infinite edges, we have a finite edge $t$ and an acute angle $\Pi(t).$

$$ \tanh t = \cos \Pi(t), \; \; \sinh t = \cot \Pi(t), $$

$$ \cosh t = \csc \Pi(t), \; \; e^{-t} = \tan \frac{ \Pi(t)}{2}. $$

Circle. Area

$$ A = 2 \pi (\cosh r - 1) = 4 \pi \sinh^2 (r/2) $$ Circumference $$ C = 2 \pi \sinh r. $$

% 523 $$ \cosh (x+y) = \cosh x \cosh y + \sinh x \sinh y $$ $$ \sinh (x+y) = \sinh x \cosh y + \cosh x \sinh y $$

% 559

If $ u = \tan (x/2), $ $$ \cos x = \frac{1 - u^2}{1 + u^2}, $$ $$ \sin x = \frac{2u}{1 + u^2}, $$ $$ \tan x = \frac{2u}{1 - u^2}, $$ $$ dx = \frac{2 du}{1 + u^2}. $$

Alright; your question, with distance scale set to $1,$ comes down naturally to four equations in four unknowns, which are angles. Area for triangles is most naturally stated in terms of angles, same on the surface of a sphere, but not in the Euclidean plane. For this material, including the distance scale $k,$ I recommend George E. Martin, The Foundations of Geometry and the Non-Euclidean Plane. For example , Theorem 32.17 on page 434 includes $k.$

Anyway, I have always worked with $k=1.$ We get four acute angles $\alpha,\beta,\gamma, \delta,$ and three (given) intrinsic distances $R,R', d,$ with the supposition that $d < R + R'.$ The four equation system becomes $$ \frac{\cos \alpha}{\sin \beta} = \frac{\cos \gamma}{\sin \delta} $$ $$ \cosh R = \cot \alpha \cot \beta $$
$$ \cosh R' = \cot \gamma \cot \delta $$ $$ \cosh d = \frac{\cos \beta}{\sin \alpha} \frac{\cos \delta}{\sin \gamma} + \sqrt{ \frac{\cos^2 \beta}{\sin^2 \alpha} -1 } \sqrt{ \frac{\cos^2 \delta}{\sin^2 \gamma} -1 } $$

After solving for the four angles, your area is $$ 2 \alpha \cosh R + 2 \gamma \cosh R' + 2 \alpha + 2 \beta + 2 \gamma + 2 \delta - 2 \pi - 2. $$


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  • $\begingroup$ It would be of help! I'm trying to learn about computing stuff in the hyperbolic plane, so every different view point is helpful to me. I do not understand what you're trying to get at with your second comment, though. I know that one can see the hyperbolic plane as a "continuous version" of a $d$-tree, if this is what you mean? I.e., that area and circumference of a circle grows exponentially, just like the number of nodes and leafs of a tree vs. its depth. $\endgroup$ – HdM Feb 23 '14 at 13:42
  • $\begingroup$ Thank you for writing it down, but I'm not sure I can follow your approach. Could you elaborate how exactly the area of the intersection relates to the 4 equation system? $\endgroup$ – HdM Feb 24 '14 at 10:36
  • $\begingroup$ I know about the laws of sine and cosine. What I don't understand is the very last step: If I see it correctly, you use the 4-equation system to compute the angles $\alpha,\beta,\gamma,\delta$. In the next line you use that to state the area. I do not understand where the very last formula comes from. Is this some sort of ratio between the areas of the two circles? I know the areas of the circles, but how can we use that to compute the intersection area? $\endgroup$ – HdM Feb 25 '14 at 10:44
  • 2
    $\begingroup$ The area of the circle with radius $R$ is $2 \pi (\cosh R - 1).$ The area of the pie slice with central angle $2 \alpha $ is $2 \alpha (\cosh R - 1).$ The are of the triangle with angle $2\alpha, \beta,\beta$ is $\pi - 2 \alpha - \2 \beta.$ The area of the lens shape, line segment as one part of the boundary, circular arc of radius $R$ the rest of the boundary, is the difference. $\endgroup$ – Will Jagy Feb 25 '14 at 17:47

I'll choose the half plane model for this. Wikipedia already tells us

A circle (curve equidistant from a central point) with center $\langle x,y\rangle$ and radius $R$ is modeled by a circle with center $\langle x,y\cosh R\rangle$ and radius $y\sinh R$.

I'll choose hyperbolic centers $(0, 1)$ and $(0, \cosh d + \sinh d)$, so I have these euclidean circles:

\begin{align*} x^2 + \bigl(y-\cosh R\bigr)^2 &= \bigl(\sinh R\bigr)^2 \\ x^2 + \bigl(y-(\cosh d + \sinh d)\cosh R'\bigr)^2 &= \bigl((\cosh d + \sinh d)\sinh R'\bigr)^2 \end{align*}

You can compute the points of intersection. This will be one hell of a symbolic expression, but nothing fundamentally complicated. Just long.

So now for the area. I'll consider treating the two circular arcs separately. For each, I'll compute the area above, since that is finite. So given an Euclidean circle with center $(0, b)$ and radius $r$, the hyperbolic area above the upper arc of that circle for $-a\le x\le a$ can be computed as

\begin{align*} \int_{-a}^a\int_{b+\sqrt{r^2-x^2}}^\infty\frac1{y^2}\,\mathrm dy\,\mathrm dx =\int_{-a}^a\frac1{b+\sqrt{r^2-x^2}}\,\mathrm dx \end{align*}

Wolfram Alpha knows the antiderivative of the integrand, but it is again a pretty large expression. Nevertheless, I see no reason why you can't plug in explicit values into that, to obtain the value of a definite integral. Similar, the area above the lower arc is given by writing $b-\sqrt{r^2-x^2}$ instead.

Then you could combine the various complicated expressions to form a single one, even more complicated. Start by taking the area above the lower arc of the upper circle, and subtract from that the area above the upper circle of the lower circle. In both cases, choose the limits so that you integrate between the points of intersection, which was our first complicated expression.

On the whole, I'd say a closed form does exist, but I wouldn't want to write it down. To give you an idea, here is some Sage code where I did the computation, and the result as well. I have no idea yet on how to obtain a simpler formula.

All of this was done assuming curvature $-1$, as is common in the half plane model. For other curvatures, you can either adapt this solution or transfer input and output of this solution by correcting for curvature.

  • $\begingroup$ Wow, thank you. This has got to be the most disgusting formula I've seen so far. I guess I have to take the physics approach here and ignore a few terms that are "small". $\endgroup$ – HdM Feb 21 '14 at 16:50
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    $\begingroup$ Note that at the moment I don't fully trust my result: for $R=3, R'=5, d=7$ it returns a significantly different result than for $R=5, R'=3, d=7$, even if computed with 1024 bit precision. So either some functions are not evaluated with that much of precision, or the formula is extremely vulnerable to numeric errors, or there is something wrong with the formula per se. $\endgroup$ – MvG Feb 22 '14 at 11:17

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