Intersection of the boundary of a convex and a circle Let $S$ be a closed, noncompact, convex subset of $\mathbb{R}^2$ which contains no lines.
Let $C$ be a circle, say centered at $(0,0)$. Is it true that $\partial S\cap C$ contains $2$ points at most when the radius of the circle becomes big enough ? (Here $\partial$ denotes the boundary) This property seems very obvious on drawings but I would prefer a formal proof.
Moreover, if we allow the convex to contain lines, then it seems that $$\text{card}(\partial S \cap C) \leq 4.$$ Any help will be much appreciated.
 A: May assume that the interior of $S$ is non-void, otherwise $S$ is contained in a line, and the statement is true. 
Take $p\in \overset{\circ}{S}$. For every $v\in S^1$, consider the half-line through $p$ in direction $v$. It will contain a closed segment ( possibly infinite). Denote by $l(v)$ the length of this segment ( may be infinite). Let now $p(v) = \frac{1}{l(v)}$. $p$ is the Minkowski functional of $(S,p)$ restricted to the circle $S^1$. It is not hard to show that $p$ is continuous.    Now, since $S$ is unbounded, there must exist $v$ so that $p(v)=0$ (otherwise $l$ would be bounded, and so $S$.  We know that $p$ is subadditive. It follows that the set of zeroes of $p$ is a closed arc of length $< \pi$ ( since $S$ does not contain a line ).
So we know we have a cone pointed at $p$ and contained in $S$ ( the cone can be of angle $0$. It's not hard to see that the same cone pointed at any other point will still be contained in $S$. ( this is called the recession cone of $S$, it's a cone of directions). Outside the arc where $l$ is infinite, the boundary of $S$ is described in polar coordinates by the continuous function $l$. Therefore, the boundary is homeomorphic to an open segment. 
May assume that the bisector of the recession cone is the vertical halfline $\mathbb{R}_{+} (0,1)$. Since the translate of any interior point of $S$ by a vector $(0, y)$ ( $y>0$) is still in $\overset{\circ}{S}$, we may assume that $p = (a, b)$, with $b > 0$. Consider the half circle opposite to $(0,1)$. The part of $S$ described around $p$ with directions in this half circle ( call is $S_{-}$ )is compact. 
Now we have two pieces of the boundary of $S$, one corresponding to directions from $p$ starting at $(0,1)$, going all the way the the right side of the recession cone, one starting at $(-1, 0)$ going all the way to the left side of the recession cone. One each of these portions, the distance from $(0,0)$ to the point on the boundary increases strictly from the point on the horizontal as the ray approaches the vertical. This is dues to the following: at each point on the boundary we have a support cone (symmetric). No line in the supporting cone can intersect the recession cone. It is not hard to show now that $l(\theta)$ will increase strictly as $\theta approaches the vertical, on each side.
The choice of $R$ is now as follows: choose $R$ so that $C((0,0), R)$ contains $S_{-}$. On each upper pieces of the boundary there will be exactly one point at distance $R$ from $(0,0)$.   
