Circle-plane relation under stereographic projection ($|z-c|=r⟺-2\Re (c)X-2\Im (c)Y+(1-|c{|}^2+r^2)Z+1+|c{|}^2-r^2=0$)

For north-pole stereographic projection $z=(X+iY)/(1-Z)$ on the unit sphere, substitute $|z{|}^2=(1+Z)/(1-Z)$ into a planar circle equation. The displayed plane equation follows, and its intersection with the sphere is a circle. The plane contains the sphere's centre exactly when $r^2=1+|c{|}^2$, giving a great circle.

A planar circle other than the unit circle lifts to a great circle exactly when it meets the unit circle at two opposite points. A great circle other than the equator intersects the equator in an antipodal pair. Conversely the plane containing a lifted circle and an antipodal pair contains their midpoint, the sphere's centre. Algebraically, adding the equations $|z-c{|}^2=r^2$ and $|-z-c{|}^2=r^2$ with $|z|=1$ gives the displayed condition.