I've been trying to find this online, in books, etc, but I can never find the expression for the metric on the unit disk
$$\mathbb{D} = \{ z \in \mathbb{C} : |z| < 1 \}$$
that has constant curvature $K < -1$. Everywhere people just do $K = -1$. I'd really appreciate the exact expression in natural cartesian coordinates. Thank you.
Things scale in a reasonable way: The metric $$ds^2 = \frac{4}{(1-x^2-y^2)^2}(dx^2 + dy^2)$$ has $K=-1$. The metric $$ds_c^2 = \frac{c^2}{(1-x^2-y^2)^2}(dx^2 + dy^2)$$ has $K=-4/c^2$.