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The sphere of centre $\mathbf{a}$ and radius $p$ has equationLet $d=|\mathbf{b}-\mathbf{a}|$ be the distance between the centres. A nondegenerate circle of intersection of two spheres exists exactly whenThe upper inequality excludes disjoint or externally tangent spheres. The lower excludes strict containment and internal tangency. It also excludes coincident centres, which give either no intersection or an entire common sphere rather than one circle. These inequalities are sufficient because a triangle with side lengths $p,q,d$ then exists with nonzero area, and rotating its third vertex about the line of centres traces a circle.

For a point on both spheres, expand their equations:Subtracting cancels ${\mathbf{r}}^2$ and gives the planeIts normal is parallel to the line of centres. Under the strict inequalities, the plane cuts either sphere in a genuine circle.