Past exam of the mathematical methods I course of the University of Oxford / 2015 / 10 / b / Solution

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.