= Solution
The quadratic-form <matrix> has <eigenvectors>
$$
e_+=\frac{(1,1,0)}{\sqrt2},\qquad e_-=\frac{(1,-1,0)}{\sqrt2},\qquad e_3=(0,0,1),
$$
with <eigenvalues> $1,9,9$. In the <principal axes> coordinates $\xi=(x_1+x_2)/\sqrt2$, $\eta=(x_1-x_2)/\sqrt2$ and $\zeta=x_3$, the equation becomes
$$
\boxed{\frac{\xi^2}{9}+\eta^2+\zeta^2=1.}
$$
It is a <prolate spheroid>, centred at the origin, with semiaxis \b[three along $e_+$ and one in every perpendicular direction]. The displayed $e_-,e_3$ are one choice of short <principal axes>.
The <sphere> has equation $\xi^2+\eta^2+\zeta^2=4$. Combining it with the ellipsoid equation gives
$$
\boxed{\xi^2=\frac{27}{8},\qquad\eta^2+\zeta^2=\frac58.}
$$
The <intersection of a prolate spheroid and a concentric sphere> therefore consists of two <circles> in planes perpendicular to $e_+$. Each has radius $\boxed{\sqrt{5/8}=\sqrt{10}/4}$, and their centres are
$$
\boxed{c_\pm=\pm\sqrt{\frac{27}{8}}\,e_+=\pm\left(\frac{3\sqrt3}{4},\frac{3\sqrt3}{4},0\right).}
$$
Their plane equations in the original coordinates are $x_1+x_2=\pm3\sqrt3/2$.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-1-q6-ellipsoid-circles.png]
{title=The ellipsoid, radius-two sphere, and their two circular intersections in principal coordinates}
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