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A negative value of either radius makes the intersection empty. Assume henceforth $\alpha ,\beta \ge 0$ and put $D=|d|$. If $D=0$, the two spheres are concentric: their intersection is the entire sphere when $\alpha =\beta >0$, the origin when $\alpha =\beta =0$, and empty otherwise.

For $D>0$, subtract their squared equations to obtain the planeThe centre of the circle of intersection of two spheres is $x_0=s\hat{d}$, whereEvery point of that plane can be written $x=x_0+u$ with $u\cdot d=0$. Its remaining equation isThus the intersection is a circle of centre $x_0$ and radius $\rho$ if $|\alpha -\beta |<D<\alpha +\beta$; a single point $x_0$ at either equality; and empty otherwise. These include tangency to a zero-radius sphere. A negative $s$ places the circle centre beyond the first sphere's centre on the side opposite $d$.

In ${\mathbb{R}}^n$ for $n\ge 2$, the same calculation gives an $(n-2)$-dimensional sphere in the perpendicular affine hyperplane when ${\rho }^2>0$, a point when ${\rho }^2=0$, and the empty set otherwise. In two dimensions the positive-radius case consists of two points. In one dimension the perpendicular hyperplane has only one point, so a nonconcentric intersection exists only when ${\rho }^2=0$. Concentric equal-radius sets coincide in every dimension.

Let $R=|\alpha |$, $S=|\beta |$ and $d=|\mathbf{b}-\mathbf{a}|$. For distinct centres, the two spherical surfaces intersect exactly whenStrict inequalities give a nondegenerate circle; equality gives a single tangency point. If $d=0$, equal radii give coincident spheres and unequal radii give no intersection, so the circle-centre formula below does not apply.

Subtracting the squared sphere equations gives the radical planeThe centre of the circle of intersection of two spheres is the projection of either sphere centre onto this plane. Write $\mathbf{c}=\mathbf{a}+\ell (\mathbf{b}-\mathbf{a})/d$; substitution gives $\ell =(d^2+{\alpha }^2-{\beta }^2)/(2d)$ and henceThe circle radius squared is $R^2-{\ell }^2$, providing another check of the intersection conditions.

Its centre equals $\mathbf{a}$ precisely when $\ell =0$, or $\boxed{{\beta }^2={\alpha }^2+d^2}$. For a positive first radius, the intersection is then a great circle of the first sphere. A triangle from its centre to an intersection point and then to the second centre is right-angled at $\mathbf{a}$, explaining the Pythagorean theorem relation geometrically.