Circle-plane relation under stereographic projection
= Circle-plane relation under stereographic projection
{title2=$|z-c|=r\ \Longleftrightarrow\ -2\Re(c)X-2\Im(c)Y+(1-|c|^2+r^2)Z+1+|c|^2-r^2=0$}
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>.