# L is a family of hyp-lines passing through a pt. Not sure how this implies rr'=(c−Re(p))c'

Lemma 1. Let $p \in \mathbb{H}$, and assume $l$ is a family of hyp-lines passing through $p$ such that $l$ is of the form $l = \{c +re^{i\theta} | 0 < θ < π\}$. For simplicity, assume the functions $c : \mathbb{R} → \mathbb{R}$ and $r : \mathbb{R} → \mathbb{R}$ are differentiable functions. Then:

(a) The left and right endpoints of l move in the same direction.

(b) Moreover, the right endpoint is moving faster if and only if the left endpoint is closer to p than the right endpoint.

The assumption that l is a Euclidean half-circle passing through p implies:

$$rr′ = (c − Re(p))c′$$

I am asked to prove the lemma by making use of the implication above.

I am not clear as to how that is implied though.

My attempt to understand:

It is clear to me that l forms a half circle in $\Bbb{H}$. I see that if the radius increases, the center has to shift. It seems that that rate at which the radius increases is proportional to the rate at which the center increases.

I'm pretty much stuck here and I don't know where to look for help.

I'd appreciate it if someone can point me in the right direction!

I drew a graph and I think I understand now:

The center is $c$, the left and right endpoints are at $(c - r)$ and $(c+ r)$ respectively, and the hyperbolic line passes through a point $p$. If you draw a line down from $p$, perpendicular to the $\Bbb{R}$-axis, and draw a line from $c$ to $p$, you form a Euclidean right triangle. The vertices are: $Re(p)$, $c$, and $p$. Using the Pythagorean Theorem, $r^2 = (c-Re(p))^2 + Im(p)^2$. Differentiate both sides to get: $2rr′ = 2(c − Re(p))(c′+0)+0$ which simplifies to $rr′ = (c − Re(p))c′$.

(Note: My class is using the book Hyperbolic Geometry by Anderson.)

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The center is $c$, the left and right endpoints are at $(c - r)$ and $(c+ r)$ respectively, and the hyperbolic line passes through a point $p$. If you draw a line down from $p$, perpendicular to the $\Bbb{R}$-axis, and draw a line from $c$ to $p$, you form a Euclidean right triangle. The vertices are: $Re(p)$, $c$, and $p$. Using the Pythagorean Theorem, $r^2 = (c-Re(p))^2 + Im(p)^2$. Differentiate both sides to get: $2rr′ = 2(c − Re(p))(c′+0)+0$ which simplifies $rr′ = (c − Re(p))c′$