Full circularity vs strong circularity for attribute grammars
You are using the terms in a weird way. A grammar can be circular or non-circular. Since FULLY testing this property is EXPTIME, one wants a simpler (easier to test) property.
This is where strong NON-circularity comes in. In strong non-circularity, you basically combine the dependencies of all possible sub-trees, even for combinations that can't appear together. So if a grammar is strong non-circular, it particularly rules out the combinations that CAN happen together. I.e. the grammar is non-circular.

So this means that strong NON-circularity implies NON-circularity. Sure you can invert implications by negating both sides of it. I think this is what you did when you say that "Full circularity implies strong circularity".
But I don't think it makes sense to call a grammar that is NOT strongly NON-circular "strong circular". IMHO that would be a confusing term, because it sounds like such a grammar was "even more circular" than other grammars, but it is the opposite - it only fails the strong-non-circularity test which isn't a necessary condition. So such a grammar would rather be "less circular" than other grammars, because it "looks circular" without necessarily being circular.
But that doesn't sound right either (because in the end, it is either circular or not). If anything, I would call it "weakly circular", because that property doesn't guarantee anything - to be certain, you still need to apply the full circularity test.

To adress the question: in practice you usually only work with S-Attribute Grammars, which don't even need to be tested for circularity, because they can't be circular.

I dug out this example (attributes starting with i are inherit attributes, attributes starting with s are synthetic ones):

S -&gt; A
    ix.1 = 1        (1)
    iy.1 = 2        (2)
    sx.0 = sz.1     (3)
A -&gt; A a
    ix.1 = sy.1     (4)
    iy.1 = ix.0     (5)
    sy.0 = sz.1     (6)
    sz.0 = 0        (7)
A -&gt; a
    sy.0 = 0        (8)
    sz.0 = ix.0     (9)
A -&gt; b
    sy.0 = iy.0     (10)
    sz.0 = 0        (11)


this grammar is non-circular, but it fails the strong-non-circularity test, because in the strong-non-circularity test, you take the union of the dependency-sets for the productions of the same nonterminal. So you have the edges for both A-&gt;a and A-&gt;b together, although they can't appear together (because you can't apply both productions at the same time).

The production of A -&gt; A a then yields the following dependency-graph:
let Q and R be the prefixes for the two A's in the production, then you have the nodes

{ Qia, Qib, Qsb, Qsc, Ria, Rib, Rsb, Rsc }


and the edges

{ (Qsb, Qia),     from (4)
  (Qia, Rib),     from (5)
  (Rib, Rsb),     from (10)
  (Rsb, Ria),     from (4)
  (Ria, Rsc),     from (9)
  (Rsc, Qsb)      from (6)
}


this is a cycle and therefore the strong-non-circularity test fails. The point why the strong-non-circularity test is wrong here, is because the edges (Rib, Rsb) and (Ria, Rsc) come from different productions.