How to get $z$ points of the two intersecting cylinders at given angle $\theta$?

Question

Cylinder at angle given by: $x^2+2y^2+z^2−2xz=2r^2$

Cylinder on $z$ axis given by: $x^2+y^2=R^2$

R is not = r.

Answer 1

I might start with:

$x=R\cos\theta\\ y = R\sin\theta$

This clearly satisfies one of the equations. Now, find $z$ in terms of $\theta$

A little reordering of the first equation. $(x-z)^2 + 2y^2 = 2R^2$

$(R\cos\theta - z)^2 = 2R^2(1-\sin^2\theta)\\ (R\cos\theta - z)^2 = 2R^2\cos^2\theta\\ R\cos\theta - z = \pm \sqrt 2 R\cos \theta\\ z = R\cos\theta (1\pm\sqrt 2)$

The set of points on the intersection are ellipses in the planes $z = x(1\pm \sqrt 2)$

Answer 2

The parametric equation of the upright cylinder is

$ x = r \cos(t) $

$ y = r \sin(t) $

Plug this into the second equation

$ r^2 \cos^2(t) + 2 r^2 \sin^2 (t) + z^2 - 2 r \cos(t) z = 2 r^2 $

And this reduces to

$ r^2 \sin^2(t) + z^2 - 2 r \cos(t) z = r^2 $

and further into

$ z^2 - 2 r\cos(t) z - r^2 \cos^2(t) = 0 $

by factoring

$ (z - r \cos(t))^2 = 2 r^2 \cos^2 (t) $

Thus

$ z = r \cos(t) (1 \pm \sqrt{2} ) $

Hence there are two seperate intersection curves between the two cylinders

The first is

$ (x, y, z) = r ( \cos(t), \sin(t) , (1 + \sqrt{2}) \cos(t) ) $

and the second is

$ (x, y, z) = r (\cos(t), \sin(t), (1 - \sqrt{2}) \cos(t) ) $

Both are ellipses.