There is a sphere equation:
$${x_1}^2 + {x_2}^2 + {x_3}^2 = R^2.$$
A quadratic surface, given by level $C$ and a set of parameters $a, e, i$, has formula:
$$a {x_1}^2 + e {x_2}^2 + i {x_3}^2 + C = 0.$$
This equation can be rewritten with the help of the diagonal matrix $M$, on the diagonal of which there are arbitrary numbers, as follows:
Elements $\ b,c,d,f,g,h$ are equal to zero.
In 3D (actually in general) what we are looking for is the smallest radius sphere centered at the origin that intersects with the quadric surface defined by a given level set from the quadratic form defined by the symmetric matrix $M$.
Is it possible to solve this problem analytically for the components of the vector $x$ using methods from differential geometry?
The radius of the smallest intersecting sphere is simply the distance from the quadric to the origin:
$$R^2 = \min_{x}\ \|x\|^2 \quad \mathrm{s.t.}\quad x^TMx = -C.$$
Assuming that $C>0$, notice that his quadratically-constrained quadratic program is of Rayleigh-quotient type, and can be rewritten as $$R^2 = \min_x\ \frac{\|x\|^2}{x^T (-M/C) x}\quad \mathrm{s.t.}\quad x^TMx < 0$$ or $$R^{-2} = \max_x\ \frac{x^T (-M/C) x}{\|x\|^2}\quad \mathrm{s.t.}\quad x^TMx < 0.$$ The solution radius will be $R^2 = -C/\lambda$, where $\lambda$ is the most-negative eigenvalue of $M$ (or no solution, if $M$ is positive-definite) and the direction $x$ of the intersection point is given by the corresponding eigenvector. When $M$ is diagonal, $\lambda$ is simply the most-negative diagonal entry.