Past exam of the mathematics course of the University of Cambridge / 2007 / ia / Paper 1 / 5B / ii

The sphere-plane intersection is governed by the orthogonal projection of the sphere's centre onto the plane. Since $\mathbf{n}$ has unit length, let $h=p-\mathbf{n}\cdot \mathbf{c}$, the signed distance from $\mathbf{c}$ to the plane along $\mathbf{n}$. The following parts determine the centre and radius and retain this sign; the radii alone cannot distinguish the two possible sides of the oriented plane.

Set ${\mathbf{b}}_0=\mathbf{c}+h\mathbf{n}$, where $h=p-\mathbf{n}\cdot \mathbf{c}$. Then $\mathbf{n}\cdot {\mathbf{b}}_0=p$, so ${\mathbf{b}}_0$ lies in the plane. For any position $\mathbf{r}$ in that plane, $\mathbf{r}-{\mathbf{b}}_0$ is perpendicular to ${\mathbf{b}}_0-\mathbf{c}=h\mathbf{n}$. The Pythagorean theorem therefore givesOn the sphere the left side is $a^2$. Thus the intersection is the circle in the plane centred at ${\mathbf{b}}_0$, with squared radius $a^2-h^2$. Its centre must therefore be $\mathbf{b}={\mathbf{b}}_0$, andIn terms of the radii, $\lambda =\pm \sqrt{a^2-{\rho }^2}$, with the sign fixed by $\lambda =p-\mathbf{n}\cdot \mathbf{c}$. In the central-plane case $\rho =a$, both signs give zero. This is the center and radius of a sphere-plane intersection formula with its orientation made explicit.

Substitute the signed displacement $\lambda =p-\mathbf{n}\cdot \mathbf{c}$ into the Pythagorean theorem relation from part (a):A genuine circle requires $|p-\mathbf{n}\cdot \mathbf{c}|<a$. Equality gives a tangent plane and a single point, interpreted as radius zero; a larger distance gives no intersection. Under the stipulated circle intersection, the radicand is positive. Its centre is $\mathbf{c}+(p-\mathbf{n}\cdot \mathbf{c})\mathbf{n}$.