I was trying to show in $\mathbb{H}$ -Poincaré half-plane model- that hypercycles defined for a given line $l$ and a given $a$ in $\mathbb{R}$ as $$H(l)=\{z \in \mathbb{H} : d_{\mathbb{H}}(z,l)=a \} $$ are not lines.
To prove something I tried to argue that given that $Mob(\mathbb{H})$ acts transitvely on the lines of $\mathbb{H}$ I can find the hypercycles of the imaginary axis and via the $\gamma \in Mob(\mathbb{H})$ that maps my line in the imaginary axis i can state $$H(l)=\gamma^{-1}(H(\gamma(l))$$
Where I'm using the fact that a Möbius transformations preserves distances. At this point I defined [already here i have doubts] $d_{\mathbb{H}}(z,l)$ for a given $z \in \mathbb{H}$ and a line $l$ as the inf in k of the distances $d_{\mathbb{H}}(z,w)$ with $w \in l$.
Given this and the general assumption $z=a+bi$ and $w=ki$ i'm left with trying to calculate the inf of $$arccosh(1+\frac{|a+(b-k)i|^{2}}{2bk})$$
EDITED FROM HERE TO SHOW MORE WORK:
given that arccosh is a monotone function i need to find the inf of $$\frac{a^{2}+{b}^{2}}{2bk}+\frac{k}{2b}-1$$
taking the derivative I'm looking to the zeroes of:
$$\frac{-2a^{2}-2b^{2}+k^{2}b}{2bk^{2}}$$
solving and remembering that i need $k>0$ i get $k=(\frac{2(a^{2}+b^{2})}{b})^{\frac{1}{2}}$
looking now for the $a+bi$ such that for a given $p$ we have: $$ d_{\mathbb{H}}(a+bi,(\frac{2(a^{2}+b^{2})}{b})^{\frac{1}{2}}i)=p $$
expanding this i got an expression of $z,z^{*}$ that is not a standard form for a line in the plane and it should end the exercise.
Is any of this correct? There is any "distance from a given point to a given line" standard formula in hyperbolic geometry? Thanks for any help and hint.
I will now answer to the questions, leaving my work in the original post to show my mistakes.
My $k$ was wrong, and the right calculations shows that the inf is taken in $$k=\sqrt{a^{2}+b^{2}}$$ Now I'm left trying do describe the locus of points in the form $a+bi$ such that for a given $p$ $$d_{\mathbb{H}}(a+bi,(\sqrt{a^{2}+b^{2}})i)=p$$ Now using the Poincare half-plane metric i get to show that $\frac{|z|}{Im z}=p$. Showing that my hypercycle is defined given an angle and it is an euclidean line passign through the origin and the given point, clearly not an hyperbolic line.