= 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):
$$
8(v^2+w^2)=5,\qquad v^2+w^2=\frac58,\qquad u^2=4-\frac58=\frac{27}{8}.
$$
Thus the intersection consists of the two <circles>
$$
\boxed{u=\pm\frac{3\sqrt6}{4},\qquad v^2+w^2=\frac58.}
$$
Their common radius is $\boxed{\sqrt{5/8}=\sqrt{10}/4}$. Transforming their centers back to the original coordinates gives
$$
\boxed{\mathbf c_\pm=\pm\left(\frac{3\sqrt3}{4},\frac{3\sqrt3}{4},0\right).}
$$
Their planes have equations $x_1+x_2=\pm3\sqrt3/2$, 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.
Back to article page
