Intersection point between circle and plane in 3d

Question

A $3D$-circle is given by the center vector, normal vector and radius.

A plane is given by a point vector and normal vector.

Find an equation for calculating the intersection points. Answer has $0$, $1$, $2$ or infinite solutions.

My linear algebra is a little rusty and I couldn't find a solution for the general case.

Under an affine transformation you may assume that the circle is in the plane $XY$ and centered at the origin. Then you just need to find the line of intersection between $\pi_C$ (the plane that contains the circle) and $\pi$ (the other plane). Once you have the line you reduced the problem to the 2D case. — mathma, Oct 10 '20 at 22:13
Yes, this is what I ended up doing. But I just projected the center of the circle as well. — Hannes Landeholm, Oct 11 '20 at 00:37
This problem is also answered here: https://math.stackexchange.com/questions/399415/3d-circle-ground-intersection — Benedikt, Jul 07 '21 at 19:28

Hannes Landeholm · Accepted Answer · 2020-10-11T00:41:37.440

So I solved this using this solution:

Compare the intersect plane with the circle plane.

If they are parallel (i.e. absolute value of dot product of normals equals 1) there are zero solutions, unless they are the same plane, in which case there are infinite solutions.

Otherwise, project* the plane in the circle plane, yielding a 2D-line. Also you may need to project the circle center in the circle plane, yielding a 2D-coordinate for the circle center, depending on where you center the coordinate system.

Then I simply solved this as a 2D line-circle intersection with the 2D center and radius.

This yields 0, 1 or 2 solutions which I projected back into 3D space (in the plane of the circle).

*Project = Take two arbitrary orthogonal axes in the plane (that are orthogonal to the normal). Each 2D dimension is the dot product of the corresponding axis and the 3D coordinate. To "reverse project", multiply each dimension with the corresponding axis and add them together. Also to move the reverse projected plane into the circle plane (restoring the information that was lost when 2D projecting), add the dot product of the circle center and the circle normal.

Intersection point between circle and plane in 3d

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