Points of $4$-contact of an ellipse and a circle Consider an ellipse $x^2 + 4y^2 = 4$ given in parametrised form $(2 \cos t, \sin t)$. At a given point $p_0 = (2 \cos t_0, \sin t_0)$ we want to measure how round the ellipse is (i.e. how similar to a circle it is). To do this, let $C(x,y) = (x-a)^2 + (y-b)^2 - \lambda$ be a circle with centre $(a,b)$. If this circle goes through the point $p_0$ then 
$$ (2\cos t_0 - a)^2 + (\sin t_0 - b)^2 -\lambda = 0 = g(t_0)$$
A point of $k$ contact between this circle and ellipse is any point for which the first $k-1$ derivatives $g^{(i)}$ vanish and $g^{(k)} \neq 0$.
For this example there are only four points for which we can have $4$-point contact. The centres of the corresponding circles are: $(\pm {3 \over 2}, 0), (0, \pm 3)$.

My question is: Why are these the only points where $4$-point contact
  is possible?

Ideally, I would like to gain geometric insight from any answer, if possible. It is clear to me that $3$-point contact is possible at every point on the ellipse.  
 A: You're correct that you can find a circle with $3$-point contact at any point $P$ on the ellipse, namely, the osculating circle at $P$. You will have $4$-point contact at the critical points $P$ of the curvature. These are the so-called vertices of the ellipse. There are, in fact, precisely 4: $(\pm 2,0)$ and $(0,\pm1)$.
EDIT: So here's the argument. Start with an arclength-parametrized plane curve $\alpha=\alpha(s)$. Recall that $T(s)=\alpha'(s)$ is the unit tangent vector and curvature $\kappa$ is defined by $T'(s)=\kappa(s)N(s)$ ($N$ is the principal normal). We're going to work with the point $\alpha(0)$ on our curve. Now consider $$f(s)=\frac12\left(\|\alpha(s)-P\|^2-r^2\right),$$ where $P=(a,b)$ is the center of the circle and $r=\|\alpha(0)-P\|$. Recall, also, that because $T(s)\cdot N(s)=0$ for all $s$, we'll have $N'(s)=-\kappa(s)T(s)$. OK, now we calculate:
\begin{align}
f'(s) &= (\alpha(s)-P)\cdot\alpha'(s) = (\alpha(s)-P)\cdot T(s); \\
f''(s) &= \kappa(s)(\alpha(s)-P)\cdot N(s) + T(s)\cdot T(s) = \kappa(s)(\alpha(s)-P)\cdot N(s)+1; \\
f'''(s) &= \kappa'(s)(\alpha(s)-P)\cdot N(s) - \kappa(s)^2 T(s)\cdot(\alpha(s)-P) + \kappa(s)N(s)\cdot T(s) \\ &= \kappa'(s)(\alpha(s)-P)\cdot N(s) - \kappa(s)^2 T(s)\cdot(\alpha(s)-P) \,.
\end{align}
Now, let's use $f'(0)=f''(0) = 0$ to conclude that $\alpha(0)-P$ is parallel to $N(0)$ and $\kappa(0) = 1/r$. This gives us the osculating circle at $P$. We then see that $f'''(0)=0$ if and only if $\kappa'(0)=0$. That is, the osculating circle will have $4$-point contact with the curve at $P$ precisely when $P$ is a critical point of $\kappa$.
