= Solution
A negative value of either radius makes the intersection empty. Assume henceforth $\alpha,\beta\geq0$ 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 plane
$$
x\cdot d=\frac{D^2+\alpha^2-\beta^2}{2}.
$$
The <centre of the circle of intersection of two spheres> is $x_0=s\widehat d$, where
$$
\widehat d=\frac dD,\qquad s=\frac{D^2+\alpha^2-\beta^2}{2D}.
$$
Every point of that plane can be written $x=x_0+u$ with $u\cdot d=0$. Its remaining equation is
$$
|u|^2=\rho^2=\alpha^2-s^2
=\frac{[(\alpha+\beta)^2-D^2][D^2-(\alpha-\beta)^2]}{4D^2}.
$$
Thus \b[the intersection is a circle] of centre $x_0$ and radius $\rho$ if $|\alpha-\beta|<D<\alpha+\beta$; \b[a single point] $x_0$ at either equality; and \b[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 $n\geq2$, 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.
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