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I have a problem with the proof of the following exercise.

Define the circle of radius $r \gt 0$ around a point c $ \in $ $H^2$ as

$C_r(c)= \; \{x \in H^2 \mid d(c,x)=r\}$ where d($\cdot$,$\cdot$) denotes the hyperbolic distance. Find the length of $C_r(c)$.

My way to solve this was as follows:

Consider the hyperboloid model. Let ($\cdot$,$\cdot$) denote the usual euclidean scalar product. Let $\langle\,\cdot,\cdot\rangle$ denote the Lorentz scalar product. Let O(2,1) denote the Lorentz Group.

It is sufficient to consider circles with center c=(0,0,1). Otherwise we apply a suitable transformation T $\in$ O(2,1). Then $C_r(c)$ can be written as $C_r(c)=H^2 \cap E$ where $E= \; \{x \in H^2 \mid \langle\,c,x\rangle=-cosh(r)\}$ is a plane.

Consider the line $L= \; \{\lambda c \in H^2 \mid \lambda \in IR\}$. Define m:=$L \cap E$ Then m=$\lambda c$ for some $\lambda \in IR$. It follows $-cosh(r)=\langle\,c,m\rangle=\langle\,c,\lambda c\rangle=\lambda \langle\,c,c\rangle=-\lambda$.

Then we get for x $\in C_r(c)$ that

$||m-x||^2=(m-x,m-x)$

$=(\lambda c-x,\lambda c-x)$

$=cosh^2(r) \cdot (c,c)-2 cosh(r) \cdot (c,x)+(x,x)$

$=cosh^2(r) \cdot \langle\,c,c\rangle-2 cosh(r) \cdot \langle\,c,x\rangle+\langle\,x,x\rangle$

$=-cosh^2(r)+2cosh^2(r)-1=cosh^2(r)-1$

$=sinh^2(r)$.

Hence the circle $C_r(c)$ can be identified with the circle defined by the equation

$||m-x||=|sinh(r)|$ (1).

Since (1) defines a euclidean circle the length must be $2\pi \cdot sinh(r)$.

While the result is correct I am in doubt whether or not my arguments hold. The problem is that this proof assumes that there is an argument that the euclidean scalar product and the Lorentz scalar product can be identified in the setup I have created. Can this somehow be proven and my proof be salvaged or should I try something completely different?

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