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The chosen unit normal points toward the plane from the origin, so the plane equation isTo locate the circle centre, project $\mathbf{p}$ onto this plane. PutThen $\mathbf{n}\cdot \mathbf{h}=d$, so $\mathbf{h}$ lies in the plane. For any other point $\mathbf{r}$ of the plane, $\mathbf{r}-\mathbf{h}$ is perpendicular to $\mathbf{n}$. The Pythagorean theorem therefore givesRestricting the sphere equation to the plane consequently gives a circle centred at $\mathbf{h}$. Thus $\mathbf{m}=\mathbf{h}$, proving that the displacement of its centre is purely normal:The same Pythagorean theorem gives $q^2={\rho }^2+{\gamma }^2$, and henceThe sign of $\gamma$ is fixed by which side of the plane contains $\mathbf{p}$; $q$ and $\rho$ alone determine only its magnitude. Substituting the signed offset yields the sphere-plane intersection radiusA genuine circle requires $|d-\mathbf{n}\cdot \mathbf{p}|<q$; equality gives tangency and $\rho =0$, while a larger distance gives no intersection. The signed distance from a point to a plane in the direction $\mathbf{n}$ is $\mathbf{n}\cdot \mathbf{p}-d=-\gamma$; the ordinary distance from a point to a plane is its absolute value.

Orient the plane $\mathbf{n}\cdot \mathbf{r}=d$ with a unit normal $\mathbf{n}$. The signed distance of a point $\mathbf{p}$ from the plane isIts absolute value is the ordinary distance from a point to a plane. The closest point in the plane is $\mathbf{p}-s\mathbf{n}$, so the displacement from the point to its projection is $-s\mathbf{n}$. Reversing the normal and $d$ reverses $s$ without changing the geometric plane or the unsigned distance. A sphere of radius $q$ centred at $\mathbf{p}$ has sphere-plane intersection radius $\sqrt{q^2-s^2}$ when $|s|<q$.