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I was trying to prove that "a hyperbolic circle in $\mathbb{D}^2$ is a Euclidean circle in $\mathbb{D}^2$ and vice-versa" ($\mathbb{D}^2$ is the Poincare's disk) then I think that I need to calculate the distance between two generics points on the Poincare's disk, but I can't find a way to do this without the omega/ideal points, because I need it for this formula:

$$d(z,z_0)=|\ln(z,z_0,z_1,z_2)|=\left|\ln\left(\dfrac{z-z_1}{z-z_2}\cdot\dfrac{z_0-z_2}{z_0-z_1}\right)\right|$$

where $z$ and $z_0$ are the points in $\mathbb{D}^2$ that I want to calculate the distance and $z_1$, $z_2$ are the generic omega/ideal points.

My idea is to calculate this distance to be the radius of the circle $$\sigma=\left\{z\in\mathbb{D}^2:d(z,z_0)=r\right\}$$

where $z_0\in\mathbb{D}^2$ is any center of the circle and $z$ is any point in $\mathbb{D}^2$ and somehow conclude that hyperbolic circle in $\mathbb{D}^2$ is a Euclidean circle in $\mathbb{D}^2$.

So, my questions are: Is there any formula to calculate generic omega/ideal points? Am I on the right way to prove that "a hyperbolic circle in $\mathbb{D}^2$ is a Euclidean circle in $\mathbb{D}^2$ and vice-versa"?

turtleman
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    The method you propose for proving this statement might or might not work, but it looks like a nightmare to me in comparison to the simplicity of the proof that I know, in which one applies the fact that every Möbius transformation of $\mathbb C^*$ takes each circle or line to a circle or line. – Lee Mosher Mar 05 '22 at 17:48
  • Yes, I tried to prove a simple statement where the center of the circle is fixed in $(0,0)$ and it worked with this method! But for any center I couldn't do anything beside ideas... can you tell me more about your known prove of this statement? – turtleman Mar 05 '22 at 17:56
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    The proof is contained in this answer that I wrote. – Lee Mosher Mar 05 '22 at 18:01
  • Thanks! And about find omega points? Is there any formula to calculate generic omega points? The only way I know to find omega points given a line in Poincare's disk is with GeoGebra... – turtleman Mar 05 '22 at 18:15
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    worth pointing out that your circle must lie strictly inside the open disc. A circle that meets the ideal boundary at a point is actually a horocycle. A cirle that meets the ideal boundary at two points is either a geodesic or an "equidistant curve" – Will Jagy Mar 05 '22 at 19:04
  • You're right, @WillJagy. With your comment, I think that maybe my way to solve the statement is wrong... and the statement just made me think about how to find omega points, I think the really intention of this post is ask if exists a formula to generic omega points... – turtleman Mar 05 '22 at 19:22

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