Properties of a eclipse on a rotated plane to see a perfect circle from the original plane view?

Lets say I have a XYZ coordinate system, with a secondary coordinate system made with rotations of 15 degrees on the X axis and 5 degrees on the Z axis. On a sketch made on this new coordinate system in the X-Z plane, what kind of eclipse would I need to make in order to see a perfect circle when looking at the sketch from the ORIGINAL X-Z plane? As in, what major and minor radii would I need, as well as any other transformations afterwards?

I've been thinking about this for a while and it seems really simple when there's a rotation on only one axis, lets say the X axis. The major radius would change on the eclipse (in the Z axis direction) and would be equal to the following equation, while the minor radius would just be the original radius of the circle since there's no "shift" in that direction.

$$major \quad radius = \frac{original \quad radius}{cos (rotation angle)}$$

I just can't wrap my head around what changes when you rotate the coordinate system around two angles. Originally I thought the second angle would only affect the minor diameter and the two components only affected their respective radii on the eclipse, but the circle doesn't match perfectly when trying this on a CAD system. Is there a subsequent rotation to the eclipse that I need to make after or some combination of trigonometry?

• The composition of two rotations is itself a rotation about some other axis. That aside, the answer to your question depends on the type of projection involved. It’s going to be rather different for a parallel (orthographic) projection than for a perspective projection, for instance. In the former case, the amount of foreshortening depends only on the angle of the ellipse’s plane to the image plane. In the latter case, the foreshortening also depends on distances ellipse from the view point and image plane. It looks like you’re using an orthogonal projection, but you need to clarify that. – amd Jan 16 '18 at 19:54
• I believe I am using an orthogonal projection, the origin of the two coordinate systems are coincident and I am using a CAD system for all of my trials. Can you expand on the extra rotation due to the composition of the axis rotations? – Richard Yang Jan 17 '18 at 14:21
• It’s not that there’s an extra rotation, but that in 3-D there’s always a single rotation that has the same effect as any composition of rotations. There’s a pretty good gloss here on Wikipedia. – amd Jan 17 '18 at 19:57