# circumference of a circle in Poincare Half Plane

by demonelite123
Tags: circle, circumference, plane, poincare
 P: 219 i am trying to figure out how to calculate the circumference of a circle in the Poincare Half Plane. I know that vertical lines are geodesics so using the arclength formula, the distance between 2 points $(x_0, y_0) and (x_1, y_1)$ on a vertical line is $ln(y_1/y_0)$. Thus, if i have a circle centered at (0, b) on the y axis with radius r, then the radius of the circle using the Poincare metric will be $\frac{1}{2}ln(\frac{b+r}{b-r})$. I want to find the circumference of the circle $x^2 + (y-b)^2 = b^2 - 1$ so i parametrize $x = \sqrt{b^2-1}cos(\theta), y = b + \sqrt{b^2-1}sin(\theta)$. I then use the arclength formula and get $C = \int^{2\pi}_0 \frac{\sqrt{b^2-1}}{b+\sqrt{b^2-1}sin(\theta)} d\theta$. I have the extra factor of $y = b + \sqrt{b^2-1}sin(\theta)$ on the bottom since I am using the Poincare metric. This evaluates to $2\pi\sqrt{b^2-1}$ which i know shouldn't be right since the circumference of this circle using the regular euclidean metric is also just $2\pi\sqrt{b^2-1}$. can someone help me figure out what went wrong here?
 P: 1,412 Why do you think your answer is wrong? Can you say more? Justify it better?

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