This is a simple question I hope, is there an easy way to solve: $$ \mathbf{x}^\mathrm{T}\mathbf{A}\mathbf{x} = c $$ for $\mathbf{x}$? (Assume $\mathbf{A}$ is positive definite).
Geometrically the solution should be an ellipse with axes given by the eigenvectors of $\mathbf{A}$ I think.
In general, you'll have more than one solution, if $c>0$. Recall that $A$ define an inner product; as particular case, for $A=I_n$, how would you solve $\lVert x\rVert_2^2=c$?
One way to do it would be to compute the square root of $A$ (well defined as $A$ is positive definite) $S$, then solve $\lVert y\rVert_2^2=c$; the solutions are of the form $y=Sx$, and it remains to solve the system ($S$ is invertible) to get $x$.