Circle of intersection of two spheres ($|p-q|<|b-a|<p+q$)

Two spheres with centres $a,b$ and positive radii $p,q$ intersect in a nondegenerate circle exactly under the displayed strict triangle inequalities. Put $d=|b-a|$. Subtracting $|r-a{|}^2=p^2$ and $|r-b{|}^2=q^2$ gives its plane,The circle centre is $a+\ell (b-a)/d$, where $\ell =(p^2-q^2+d^2)/(2d)$, and its squared radius is $p^2-{\ell }^2$. This is positive exactly under the strict inequalities. Equality gives internal or external tangency; coincident equal spheres have an entire sphere in common rather than a unique intersection circle.