Anyway, a circle in space would be the intersection of a plane and a (hollow) sphere. The sphere equation is easy. Work out the plane equation from the direction cosines (I don't even remember what a direction cosine is!) and the circle center point, substitute one equation in the other.
Take the crossproduct of the unit vector cosA,cosB,cosC,(A,B,C the direction angles) with the vector x-x1,y-y1,z-z1. ;then take the self dot product and set it equal to R^2.
For the crossproduct, I get [cosB(z-z1)-cosC(y-y1)],[cosC(x-x1)-cosA(z-z1)],[cosA(y-y1)-cosB(x-x1)]
This vector is in the direction parallel to the plane of the circle and has a magnitude of R. Taking the self dot product equal to R^2 should yield the equation, namely
zekeman
a single equation in 3D necessarily represents a surface. I didn't check your way of reasoning, but your equation should be a cylinder: you still need to intersect it with a plane to get a circle (hence again a system of two equations).
Prex,
You are correct. Mine is the eq of the cylinder perpendicular to the circle in space and, indeed I would need the eq of the intersecting plane to complete the answer.
However, a discussion of the more general closed planer curve problem rotated in space, namely
F(x,y)=0
would be more interesting, since it would not involve a familiar intersection of a plane and a sphere or cylinder.
Try it in your spare time!
