Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2023 / Paper 1 / 6 / c / Solution

The orthogonal change of coordinates preserves the sphere equation, giving $u^2+v^2+w^2=4$. Subtract it from the spheroid equation in part (b):Thus the intersection consists of the two circlesTheir common radius is $\boxed{\sqrt{5/8}=\sqrt{10}/4}$. Transforming their centers back to the original coordinates givesTheir planes have equations $x_1+x_2=\pm 3\sqrt{3}/2$, each with normal vector $(1,1,0)$. The difference ${\mathbf{c}}_{+}-{\mathbf{c}}_{-}$ is parallel to this normal, proving that the line joining the centers is perpendicular to both planes. This is the intersection of a prolate spheroid and a concentric sphere in the regime between the short and long semiaxes.