Intersection of Hyperbolic Plane with affine plane $\{t = \cosh r \}$

Question

Why does the intersection of the hiperbolic plane with the affine plane $\{t = \cosh r \}$ give a center $(\cosh r, 0, 0)$ in the euclidean space?

Answer 1

I'm assuming your model of hyperbolic space is the hyperboloid model

$$ \mathcal H^2 = \{(t, x, y) \in \mathbb R^3 : t > 0,~ t^2 = x^2 + y^2 + 1\}$$ with the usual inner product.

Note that $\mathcal H^2$ as a subset of $\mathbb R^3$ is rotationally symmetric about the $t$-axis, and the plane $p := \{(t, x, y) : t = \cosh r\}$ is also rotationally symmetric about this axis. Since the circle we are looking at is the intersection of these two sets, it must also be symmetric about this axis.

The center in $\mathbb R^3$ of this circle must now lie on both the plane $p$ and the axis of symmetry. The plane and the axis intersect in a single point, which is $(\cosh r, 0, 0)$, so the midpoint of the circle can be no other point.