Orthogonal projection of a circle between planes ($b/a=|{\mathbf{n}}_1\cdot {\mathbf{n}}_2|$)

Under orthogonal projection from one plane onto another, a circle of radius $R$ becomes an ellipse with semiaxes $R$ and $R|\cos \alpha |$, where $\alpha$ is the angle between the planes' unit normal vectors. To prove this, choose one in-plane unit vector parallel to the planes' intersection and the other perpendicular to it. The first vector is unchanged by the projection; the second has projected length $|\cos \alpha |$ and remains perpendicular to the first. Parametrizing the circle in that orthonormal basis therefore gives projected coordinates $(R\cos t,R\cos \alpha \sin t)$. Parallel planes have a circular image, and perpendicular planes give the degenerate case of a line segment. An arbitrary center simply projects to the center of the image.