# Formula of a circle in 3D plane to equirectangular 2D plane

I have a plane $P$ that intersects a sphere $S$ with a radius $R$. The result of this intersection is a circle $C$ with the distance between its center and the sphere's center is $s$. In my case is always $R \geqslant s$ so I get a circle or a point-circle as a result of this intersection.

Edit \begin{align*} P: && ux + vy + wz + d =0\\ S: && x^2 + y^2 + z^2 = R^2 \end{align*} The formula of $C$ is of the form: $$Ax^2 + Bxy + Cy^2 + Dx + Ey + F =0.$$

What is the formula to represent (equirectangular project) $C$ on a 2D plane ($xOy$ plane)? I would have the circle $C$ plotted so on a 2D plane (it would like look the day-night terminator as in map below: Globally it looks like: Thanks for any help :)

• Your formula for $C$ looks like a sphere to me.
– mvw
Jun 17 '18 at 13:19
• Sorry, you are right, I edited it Jun 17 '18 at 13:25
• My goal is to represent a circle (not necesserily a great-circle) around a sphere (e.g. the globe) on a 2d plane (e.g. cylindrical equirectangular projection) Jun 17 '18 at 13:28
• The you need to put all points from the circle through your projection.
– mvw
Jun 17 '18 at 13:46