Undecidable unary languages with linear-size circuits ($U_A=\{1^n:n\in A\}$)

For any set of lengths $A$, choose at length $n$ a constant-zero Boolean circuit when $n\notin A$, and an AND of all inputs when $n\in A$. This uses $O(n+1)$ gates. If $A$ is undecidable, so is $U_A$, yet it belongs to P/poly. Since all NP languages are decidable by certificate enumeration, this proves P/poly is not equal to NP; it does not resolve containment in the other direction.