Is the polar dual of a semi-algebraic convex body also semi-algebraic? Call a convex body $C\subset\Bbb R^n$ semi-algebraic if it can be written as
$$(*)\quad C=\bigcap_{i\in I}\, \{x\in \Bbb R^n\mid p_i(x)\le 0\}$$
with polynomials $p_i\in\Bbb R[X_1,...,X_n]$ and a finite index set $I$.

Question: Is the polar dual $C^\circ$ of a semi-algebraic convex body again semi-algebraic?

Where polar dual means
$$C^\circ := \{y\in\Bbb R^n\mid \langle x,y\rangle\le 1\text{ for all $x\in\Bbb R^n$}\}.$$
 A: The comment of Robert brought me onto the right track.
Say $I=\{1,...,m\}$, then
\begin{align}
\Bbb R^n\setminus C^\circ 
&= \{\,y\in\Bbb R^n\mid \exists x\in C\colon\langle x,y\rangle >1\,\}
\\&= \{\,y\in\Bbb R^n\mid \exists x\in\Bbb R^n\colon p_1(x)\le 0 \land\cdots\land p_m(x)\le0 \land \langle x,y\rangle>1\,\}
\\&=\pi\, \{(x,y)\in\Bbb R^n\times\Bbb R^n\mid p_1(x)\le 0 \land\cdots\land p_m(x)\le0 \land \langle x,y\rangle>1\,\},
\end{align}
where $\pi$ is the projection $(x,y)\mapsto y$.
So $\Bbb R^n\setminus C^\circ$ is the projection of a semi-algebraic set. By Tarski-Seidenberg $\Bbb R^n\setminus C^\circ$ is a semi-algebraic set, and so is its complement $C^\circ$.
What I don't yet see is why it is necessarily of the form $(*)$, i.e. using only intersections. In fact, I think the following pair of shapes (which are polar duals of each other) forms a counterexample:

The shape on the left is semi-algebraic convex in my sense, being the intersection of a disc and two halfspaces.
However, I can't see how the shape on the right can be written as an intersection of algebraic sets.
