I have a problem of the form
$$\begin{equation*}
\begin{aligned}
& \underset{x}{\text{minimize}}
& & \lVert \mathbf{x-a} \rVert_2 \\
& \text{subject to}
& & \lVert \mathbf{x-v} \rVert_2 \leq \lVert \mathbf{s-v} \rVert_2
\end{aligned}
\end{equation*}$$
where $\mathbf{a}$, $\mathbf{s}$, $\mathbf{v}$ are constant. I am not sure how I could reformulate this into the standard form with $Ax \leq b$ inequality constraints and would appreciate any help. I have seen some examples for L1 norm (e.g. here) but not sure how to apply it in this case.
Best Answer
The feasible set is not a polyhedral, hence it can't be directly be described in terms of $Ax \le b$. Unless perhaps you change the feasible set of which the optimal solution is still preserved.
Your problem however can be solved.
First, we can see if $a$ is in the feasible set. If it is, then $x=a$ is the solution.
Suppose $a$ is not in the feasible set. Draw a straight line between the center, $v$ and $a$. The closest point on the sphere that is the closest to $a$ must be on the surface and on the straight line. $$x^*=v+\frac{(a-v)}{\|a-v\|}\|s-v\|$$