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I was making some routine calculations when I found the strange following result : in the hyperbolic plane (with the Gauss curvature set at K=-1), let's $C$ a circle of center $O$ and radius $R$ and $A$ a point with $OA=R+h$, $AT$ the tangent (at $T$) to the circle, and $A_0$ the intersection of $OA$ and $C$. In Euclidean geometry, we have $AT=\sqrt {2hR+h^2}$, which tends towards infinity as $R$ does (in other words, applied to a sphere, the horizon expands as the radius of the sphere augments and the sphere flattens), but in hyperbolic geometry, the triangle $ATO$ is rectangle in $T$, so $\cosh OA=\cosh AT\cosh OT$, $\cosh AT=\frac{\cosh (R+h)}{\cosh R}\simeq e^h$ (well, there is a small error getting very quickly to 0), which means that even at the limit, the horizon has no relation to the angle of parallelism. A confirmation is given by calculation of angles : the angle $\theta$ of parallelism (i.e. the angle between $OA$ and $D$, a parallel asymptote through $A$ to the tangent at the circle in $A_0$) is given by the Lobatchevski formula $\sin\theta=\frac{1 }{\cosh h}$, while the angle $\alpha=\widehat{OAT}$ satisfies (sinus law) $\sin\alpha=\frac{\sinh R }{\sinh (R+ h)}$, which tends very fast to $\frac{1 }{e^ h}$, a difference of a factor of 2... What is happening here ?

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    $\begingroup$ I couldn't follow the stuff about "angle of parallelism": could you clarify which "circle in $A_0$" you use to define $D$? Up until that point your calculations seem fine, but I'm unsure in what sense they are "paradoxical". It is certainly true that the hyperbolic plane behaves very differently "at infinity" from the Euclidean plane. $\endgroup$
    – HJRW
    May 13, 2021 at 12:21

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Sorry , I found the answer by myself : the limit of the circle as $R$ tends to infinity is NOT the tangent (at $A_0$) ! To see it, fix $B$ at a distance $d$ from $A_0$ on the tangent, and call $B_0$ the point on the circle and on $OB$. It is easy to check that $\cosh d = \frac{\cosh (R+B_0B)}{\cosh R}\simeq \exp(B_0B)$... Of course, I could also have noticed that the limit is not a line, but the horocycle having its center at the point at infinity of $AO$, and passing through $A_0$...

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  • $\begingroup$ ...while in spherical geometry, for each $h$ there is an $R_h$ (smaller than the geodesic radius) such that for all $R>R_h$ and all directions, $A_0A<h$. Moreover, the closer $R$ gets to the geodesic radius the smaller is $\max(A_0A)$ $\endgroup$ May 13, 2021 at 21:19

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