Circle-Circle intersection area in the hyperbolic space

Question

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).

Answer 1

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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Answer 2

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.