 Center and radius of a sphere-plane intersection

For a sphere with centre $c$ and radius $a$, and a plane $n\cdot r=p$ with unit normal $n$, the orthogonal projection of $c$ onto the plane is $b=c+(p-n\cdot c)n$. The orthogonal decomposition $r-c=(r-b)+(b-c)$ gives the intersection circle radius by the Pythagorean theorem. A genuine circle requires $|p-n\cdot c|<a$; equality gives tangency and larger distance gives no intersection. The signed displacement is not determined by the two radii alone: it is $\pm \sqrt{a^2-{\rho }^2}$ with sign fixed by the oriented plane.