How does the angle of parallelism relate to the arc of a circle and a point outside?

In Hyperbolic Geometry, I'm trying to figure out what happens to the "visibility" of a circle when a point outside increases in distance away from the circle.

So given a point p outside the circle, there exists two tangents from the circle that cross point p. As p moves farther away these tangents move and the arc between them increases. I'm given the equations for the angle of parallelism(2arcTan(e^-x)) and the length of the circle (2piSinH(r)).

I know the angle goes to 0 as x->infinity.

I would prefer the answer to finding an expression to finding the "visibility" of the circle arc or at least a hint in the right direction to be able to know where to assume the angle of parallelism lies in relation to a point outside a circle.


Consider a circle $c$ centered at point $C$ and consider $P$ a point outside the circle as shown below. (I am using here the Klein model.)

Let $\lambda$ be the angle of visibility of the circle if it is seen from $P$.

The first observation is that if $P$ goes to the infinity then the angle of visibility goes to zero just like in Euclidean geometry. That is, the farther we are from the circle, the smaller it looks. This is not surprising.

Second, note that the angles at the tangent points are right angles. If, again, $P$ goes to the infinity then $\gamma$ tends to $\Pi(r)$, the angle of parallelism belonging to the radius of our circle.

While in the Euclidean geometry $2\gamma$ would tend to $\pi$, in Hyperbolic geometry $2\gamma$ tends to $2\Pi(r)$ as the $P$ back away from the center of circle.

Then: the arc lenghth is proportional to the angle at the center.

I don't know if this answers the question.

  • $\begingroup$ One interesting point here is there is a limit arc length. I.e., there is an absolute length such that you will always see less than this arc length, regardless on radius of circle and the point you're viewing it from. (To see this, consider a horocycle seen from the infinity -- even in this extreme case, the arc length is finite.) $\endgroup$ – Marek14 Aug 8 '17 at 8:51

Just an first long comment to get the question clear, will be updated later.

I am just puzzeling along, but I guess this will be to long for a comment.

Lets first work out the Euclidian case (then at least we know where we are talking about)

So (at the same time filling in some gaps)

We have a circle $C$ with centre $O$ with some fixed radius $r$ and a point $P$ outside this circle

From $P$ we have two tangents $t_1$ and $t_2$ to circle $C$.

Lets call the points where $t_1$ and $t_2$ touch $C$ , $T_1$ and $T_2$ and then the question is what is the arc of visibility (the arc of $C$ facing between $T_1$ and $T_2$ facing $P$ ?

and how long is it as function of $\angle T_1PT_2$ (do I have this all correct?)

A bit geometry learns us that that arc has the length $r (\pi -\angle T_1PT_2 ) $ (yes radians please) Just because all we know about the kite $OT_1PT_2$ , $\angle OT_1P$ and $\angle OT_2P$ (see https://en.wikipedia.org/wiki/Kite_%28geometry%29 )

then the question is what is the same then in hyperbolic geometry.

So what is the same? $\angle OT_1P$ and $\angle OT_2P$ are still $\pi/2$ radians.

what is different? - thats more for the next edit.

please update the question accordingly and we can continue the puzzle :)


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