The sphere $x^2 + y^2 + z^2 = 1$ and the ellipsoid $a x^2 + y^2 + b z^2 = 1$ are given where $b > 1 > a > 0$. We compute their intersection. It turns out that these are the points $$r_{\pm\pm} = \left(x, \pm\sqrt{1 - \alpha x^2}, \pm\sqrt{\beta}x\right)$$ where $\alpha = \frac{b - a}{b - 1}$, $\beta = \frac{1 - a}{b - 1}$ (hence $1 +\beta = \alpha$), for $|x|\leq 1/\sqrt{\alpha}$. It turns out that these points lie on a two circles (e.g., the first is union of $r_{++}$ and $r_{-+}$ types of points), which becomes aparent if we rotate each of the circles around $y$-axis, so that it lies in the $yz$-plane (for each circle we need different rotation).
Is there any obvious reason why the intersection consist of circles and not two ellipses (as one would naively expect) or even worse - some non-planar curves?
Well, it's hard to distinguish what's obvious or not, but here's how you can think of it given a couple of preliminaries as obvious:
If you believe that the intersection has two connected components, and if you think that it's obvious that each of these lies on a distinguished hyperplane (affine subspace) of $\mathbb {R}^3$, then you can easily see that each must indeed be a circle since the intersection of a sphere with a hyperplane is always a circle (unless it is either empty or only a single points).
Put differently, how would you embed an ellipse in a sphere? You'd probably still pick a hyperplane on which the ellipse lies, right? But then the ellipse is automatically a circle.