= Solution
Let $R=|\alpha|$, $S=|\beta|$ and $d=|\mathbf b-\mathbf a|$. For distinct centres, the two spherical surfaces intersect exactly when
$$
\boxed{|R-S|\le d\le R+S.}
$$
Strict 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 plane>
$$
2(\mathbf b-\mathbf a)\cdot\mathbf r
=|\mathbf b|^2-|\mathbf a|^2+\alpha^2-\beta^2.
$$
The <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 hence
$$
\boxed{\mathbf c=\frac12\left(\mathbf a+\mathbf b+
\frac{(\alpha^2-\beta^2)(\mathbf b-\mathbf a)}{|\mathbf b-\mathbf a|^2}\right).}
$$
The 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.
Back to article page
