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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.

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.

The center of the sphere-plane intersection is the perpendicular projection of the sphere's center onto the plane:The distance from $\mathbf{c}$ to the plane is $3<4$, so the intersection is a genuine circle. A right triangle with hypotenuse equal to the sphere radius givesIts plane is the given plane; displacements within it are perpendicular to $\mathbf{n}$.

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

For a general, not necessarily unit, normal vector ${\mathbf{n}}_0$ in ${\mathbf{n}}_0\cdot \mathbf{x}=d_0$, the oriented distance is $({\mathbf{n}}_0\cdot \mathbf{p}-d_0)/\Vert {\mathbf{n}}_0\Vert$. In the worked example of distance from a point to a plane, orientation toward $(2,-1,2)$ assigns $+8/3$ to $(4,1,2)$; reversing that unit normal assigns $-8/3$. The closest point on a plane and the unsigned distance are unaffected.