# Equation of a circle in 3D space

• Aug 13th 2008, 03:47 AM
mathsrex
Equation of a circle in 3D space
Hi All,

Thanks for answering my last question about circle in 2D. I got a question about a circle in 3D space here.

There is a sphere at the origin (0,0,0) with a radius of 6. A point at (10, 12, 15) project a cone shape on the sphere where the cone has a cone angle of 10 degree. The concept is similar to the satellite project signal on earth.

So my question is: What is the equation of the circle which is the intersection of the sphere and cone? I know there will be two circles. But I only want the one closer to the point. Thank you!
• Aug 13th 2008, 01:53 PM
TKHunny
One might consider working a slightly simpler example, first.

Rather than (10,12,15), why not try it at (0,0,21.65640783).

With a 10º vertex angle, this gives a nice cone:

$\frac{(z-21.65640783)^2}{21.65640783^{2}}\;=\;\frac{x^{2}+y ^{2}}{1.89469018}$

With your 6-radius sphere, this is easily solved for the intersection, giving first z = 5.838274311 and defining the circle nicely with radius 1.38367376.

Your remaining task is to rotate the result to where you want it. Alternatively, one could rotate where you want it to the easier solution.