A general case of circular logic is when you have an assumption $P$, and $Q$ serves as ‘proof’ for $P$, but the only reason $Q$ is said to be true is because of $P$. In other words, the proof for $P$ is only valid assuming the truth of $P$.
I would symbolize this as: $(P\implies Q)\implies P$
Is there a way of checking if this is a tautology or contradiction, etc.?
Circular logic itself should work out to a tautology, yes, because it shows nothing at all. For example, if you prove $P$ by assuming that $P$, you effectively show that $P \to P$ is true ... but $P \to P$ is a tautology, and so effectively you have shown nothing at all.
With your example, assuming $P$ to prove $Q$, as well as using $Q$ to prove $P$, you actually are showing a bit more than nothing, for you are showing $P \to Q$ as well as $Q \to P$. But of course, as a proof of $P$, you effectively still have shown nothing more than $P \to P$, which is to say: nothing at all.
OK, so then where is the mistake in circular logic? Well, one way to think about it is if you take the circular proof as a whole to prove that $P$. That is, it is when you are trying to infer $P$ from something that really shows nothing more than $P \to P$. In logic, then, you'd effectively argue that:
$(P \to P) \to P$
But if $P$ is false, then this statement is false. And so, $P$ does not follow from $P \to P$. Indeed, nothing that is not a tautology follows from $P \to P$
And again, applied to your case. OK, so we show $P \to Q$ well as $Q \to P$. Can we conclude $P$ from that? That is, can we say that it is always the case that:
$((P \to Q) \land (Q \to P)) \to P$
No, clearly not: set both $P$ and $Q$ to false, and this statement is false as well. Hence, showing that $Q$ follow from $P$ and vice versa does not show that $P$, period. That's the mistake.