The set consists of all points in a given circle is an algebraic set? Let $D=\left\{{(x,y)|x^2+y^2\leq a^2}\right\}$ where $x, y\in \mathbb R$.  
Is it true that $D$ is an algebraic set ? 
I guess that's maybe a silly question but I can't prove or disprove it.

A subset $U\subset \mathbb{R}^n$ with $U\neq \mathbb{R}$ whose interior is non-empty (for the topology induced by the euclidean metric) can never be an algebraic set. Polynomials that vanish on an open set, are constantly zero. If you assume $U$ to be an algebraic set, then by definition it would be the set of common zeros of some set of polynomials, but then all these polynomials must be constantly zero and it must be $U=V(\{0\})=\mathbb{R}^n$, contradicting our assumption.
Your example of the open disk is a special case of this. As long as $a\neq 0$, $D$ is subset of $\mathbb{R}^2$ (with $D\neq\mathbb{R}^2$) that has a non-empty interior, so it cannot be an algebraic set.
