A query about Complexity Class $NC$
The complexity class $\mathsf{NC}$ is a uniform class. There are constant size circuits which don't belong to $\mathsf{NC}$.

What does that mean? Most circuit complexity classes don't really make sense for individual circuits. A single circuit can't have polynomial size, or rather, this doesn't mean anything (every circuit has size at most $n^k$ for some $k$). Instead, when talking about a circuit belong to a complexity class, we are really talking about sequences of circuits. These sequences can be uniform – that is, can be generated by a Turing machine – or arbitrary. (In the uniform case, you can get different complexity classes by constraining the strength of the Turing machine.)

The complexity class $\mathsf{NC}$ consists of uniform circuits of polynomial size and polylogarithmic depth. Consider the following sequence of circuits: $C_n$ (with $n$ inputs) is either the constant 0 or the constant 1, depending on whether the $n$th Turing machine halts on the empty input. This sequence doesn't belong to $\mathsf{NC}$ (since it's not even computable), but all circuits have constant size.

In order to compare $\mathsf{NC}$ and polynomial size circuits, you have two options. Either you compare $\mathsf{NC}$ to $\mathsf{P}$, which is the uniform analog of polynomial size circuits, or you compare non-uniform $\mathsf{NC}$ to polynomial size circuits (a complexity class known as $\mathsf{P/poly}$). In both cases, it is conjectured that the two classes are different, but this hasn't been proved yet.

One way of separating $\mathsf{NC}$ from $\mathsf{P}$ would be to separate non-uniform $\mathsf{NC}$ from $\mathsf{P}$, that is, to give an example of a problem in $\mathsf{P}$ (say, some $\mathsf{P}$-complete problem) that cannot be solved using $\mathsf{NC}$-circuits. In the same way, you could prove that $\mathsf{P} \neq \mathsf{NP}$ by showing that $\mathsf{SAT}$ doesn't have polynomial size circuits. In both cases, you are actually trying to prove stronger claims, which are believed to hold.