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.

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 :)