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The title of this question might seem not so clear, as straight lines are by definition curves of zero curvature; however, i refer to lines in Poincare unit disk model, in which straight lines are modeled as portions of circles orthogonal to the boundary of the unit disk (i.e circle that intersects the unit circle at right angles).

Therefore, my question is: given the coordinates of two points in the Poincare disk, what is the curvature of the orthogonal circle connecting them? I prefer more conceptually transparent ways to arrive at the formula for curvature.

user2554
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  • Under inversion in the boundary circle of the Poincare disk, the circles which are fixed are exactly the Poincare lines. Thus, given two points in the disk, invert them to two points outside the disk. The four points determine a unique circle. – Somos Aug 30 '20 at 20:28
  • Your question is somewhat unclear. In the Poincare disc model, the formula for the curvature of the portions of circles orthogonal to the boundary of the unit disk is very simple, it is the constant $0$. That's why they are the straight lines of hyperbolic geometry. Are you asking how that formula is derived? – Lee Mosher Aug 31 '20 at 13:57
  • I meant what is the curvature of this orthogonal circle literally, i.e $1/r$ where $r$ is the radius of this orthogonal circle. I know that this is the way to model straight lines with zero geodesic curvature, but i wanted to simply look at the circular arc connecting two points and and calculate it's radius when it viewed euclidly, without any sophistications. Anyway, eventually i found the answer (initially i thought i won't succeed alone...). – user2554 Aug 31 '20 at 14:05

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