Intersection curve parametrization

Question

How can we find parametrization of the intersection curve between a pseudo-sphere and sphere ?

Elimination of variables appears not readily amenable/elegant. Thanks for the same.

Answer 1

The $z \gt 0$ half of a pseudosphere with radius $r$ is a surface of revolution (rotated around the $z$ axis) described by $$\left\lbrace ~ \begin{aligned} x(t) &= r t - r \tanh t \\ y(t) &= r \operatorname{sech} t = \frac{r}{\cosh t} \\ \end{aligned} \right.$$ where $0 \lt t \in \mathbb{R}$.

This intersects a circle of radius $r$ centered at origin when $$x(t)^2 + y(t)^2 = r^2$$ i.e. $$(r t - r \tanh t)^2 + \frac{r^2}{(\cosh t)^2} - r^2 = 0$$ The degenerate sphere case ($r = 0$) is uninteresting, so we can assume $r \gt 0$. For $t \gt 0$, $\cosh t \gt 0$, so we can divide the equation by $t r^2$ and multiply by $\cosh t$, getting $$\begin{aligned} t \cosh t - 2 \sinh t &= 0 \\ t \cosh t &= 2 \sinh t \\ t &= 2 \tanh t \\ \end{aligned}$$ which has only one solution for $0 \lt t \in \mathbb{R}$, $$t \approx 1.91500804815453748135300306100481565057336256878593...$$

Because both shapes are symmetric around the $z$ axis, the intersection is a circle of radius $y(t)$ perpendicular to the $z$ axis at $z = x(t)$, i.e. $$\left\lbrace ~ \begin{aligned} R = y(t) &\approx 0.28841990894499 \, r \\ z = x(t) &\approx 0.95750402407727 \, r \\ \end{aligned}\right.$$

Answer 2

Three circles having center on $z$ axis at about $C_1(0,0,-0.95936);\;C_2(0,0,0);\;C_3(0,0,-0.95936)$ and radius about $r_1=r_3=0.28786$ and $r_2=1$

This is the best I could do, since are involved transcendent function an exact solution is impossible to get.

Hope it helps