Osculating circle (This question may be too elementary for this site — I'm fine if it needs to be moved to math.stackexchange.)
If I approximate a nice planar curve by a straight line, the tangent, then the second derivative tells me which side of the line the curve lies on locally.  If instead I approximate the curve by a circle — the osculating circle — what determines the answer to the analogous question?  Which side of the osculating circle does the curve lie on locally?

Update:  I'm accepting one of the answers below, but it's surprising and with my poor geometric intuition not particularly easy.  This situation is the exact opposite of the tangent line analogy I made in the question.  In that case the graph of a function lies locally on one side or the other of the tangent line at generic points, and only crosses at inflection points.  But for the osculating circle it is the reverse: the curve crosses the circle at generic points, and only lies (locally) inside or outside the circle at maxima or minima of the curvature.
The can be deduced from the Tait–Kneser theorem, as the answer below claims.  A nice explication by Ghys, Tabachnikov and Timorin can be found at Osculating curves: around the Tait–Kneser Theorem.
The theorem itself says wherever the curvature is monotonic (e.g. when parametrized by arclength), the osculating circles are pairwise disjoint and nested.  It takes some geometric visualization to realize this means the circles can not lie on one side of the curve as the point of contact moves.
This can be a challenge to see on actual graphics, because the curve and circle agree up to second order terms.  A Mathematica demonstration may be found here which allows one to select in particular the example of an ellipse which was mentioned in the comments.
 A: Tait–Kneser theorem says that generically curves crosses its osculating circle. If not (that is, if the curve is locally supported by its osculating circle), then the point is a vertex of the curve, but the supporting condition is stronger a bit.
If it is a point of local minimim/maximum of the curvature, then the curve lies locally outside/inside of its osculating circle [see 6.4 here].
A: In retrospect I realize one does not need the full strength of Tait-Kneser.  The curve and the circle agree up to quadratic order.  So the error, the difference between the two, looks like a constant times $s^3+O(s^4)$ when parametrized by arclength $s$ measured from the point of contact.  The error has to change sign when $s=0$.
A: Assuming the curve $\gamma(s)$ is parametrised in arc length, if for a disk $B(p,r)\subset\mathbb R^2$ one has $\gamma(s_0)\in\partial B(p,r)\subset\mathbb R^2$, then $\gamma(s)$ is locally outside the disk $B(p,r)$ at $s_0$ iff the distance $\|\gamma(s)-p\|^2$ has a local minimum at $s=s_0$. For the osculating circle at $\gamma(s_0)$, that is $p=\gamma(s_0)+\frac{\ddot\gamma(s_0)}{\|\ddot\gamma(s_0)\|^2}$ and $r=\frac{1}{\|\ddot\gamma(s_0)\|}$, we get $\|\gamma(s)-p\|^2$ has a degenerate stationary point at$s=s_0$, and a necessary (resp.  sufficient) condition for being a local minimum is $\partial^3_s {\|\gamma(s)-p\|^2}\big|_{s=s_0}=0$ and $\partial^4_s {\|\gamma(s)-p\|^2}\big|_{s=s_0}\ge0$ (resp., with strict inequality), that is $\ddot\gamma\cdot\dddot\gamma=0$ and $\dot\gamma\cdot\dddot\gamma+\ddot\gamma\cdot\ddddot\gamma\ge0$ (Resp. $>0$)
