Such a coloring does exist : Let $c_1,c_2,...,c_{2000}$ be the colors. In th following, subscripts are considered modulo 2000. Let's give an orthonormal system of coordinates. Now, for any given real number $a$, let $D_a$ be the line with equation $x=a$. Let $a$ be a decimal number, which means that its decimal expansion has a finite number of non-zero digits after the decimal point. Let $n = n(a)$ be the rank of its last nonzero digit, which will be called the length of $a$ Then color all the line $D_a$ with color $c_n$. We then color each point of the plane with decimal x-coordinate. Now, give any of the colors to all the other points to construct a coloring of the entire plane. Now, let $C$ be a circle with center $M(x_M,y_M)$ and radius $r$. Since the interval $[x_M - r ; x_M + r]$ contains decimal points with all lengths sufficientely large, we deduce that for each $i \in \{1,2,...,2000 \}$ the circle $C$ intersect a line $D_a$ which has been colored with $c_i$, which proves the desired property of the coloring. Pierre.

Yes, the plane can be coloured in the required manner. Label the colours $C_i, 1 \leq i \leq n$. Colour the lines $L_i = \{y= \frac{a_i}{b_i}: a_i, b_i \in Z, b_i == i (mod n)\}$ colour $C_i$, and the lines $y=r$, $r$ is 0 or irrational, arbitrarily. Lemma: There exist $a,b \in Z$ such that for any $L<U$, $L,U \in R$, $L <\frac{a}{nb+i}<U$. Proof: The difference $U(nb+i)-L(nb+i)=(U-L)(nb+i)$ can be made larger than 1 since $b$ can be made arbitrarily large. Therefore, there exists $a \in Z$ such that $U(nb+i)>a>L(nb+i)$ or equivalently, $U>\frac{a}{nb+i}>L$, proving the lemma. It follows, by the lemma, that any circle with maximum $y$ coordinate $U$ and minimum $y$ coordinate $L$ intersects some member of $L_i$ which, by construction, is coloured $C_i$, proving that every circle contains a point coloured $C_i$ for all $1 \leq i \leq n$.

Sorry for reviving this very, very old topic, but I'm struck by it. We want to show the points of the plane can be colored with $ n$ colors, so that any non-degenerate circle contains points of all colors. 1. The set $ D = \mathbb{Q}^2$ is dense in $ \mathbb{R}^2$; 2. There exist $ n$ real numbers $ x_i$, linearly independent over $ \mathbb{Q}$ (take them from a Hamel basis); 3. The sets $ D_i = (x_i, x_i) + D$ are disjoint, and each still dense in $ \mathbb{R}^2$; 4. Color $ D_i$ using color $ C_i$, and the rest of the points of the plane (randomly) at will. So why all the fuss?

There exist circles that do not pass through any point of $ {\bf Q}^2$ (e.g., center $ (0, 0)$, radius $ \pi$), and translates of these circles avoid some of your $ D_i$. (Actually, we can probably avoid all of your $ D_i$ simultaneously.) Density is not sufficient, as a circle does not contain an open set in $ {\bf R}^2$.

Ah, the old confusion between circle and disk (I probably got confused by the word "contain", as it seemed to point to the interior of the circle (i.e. disk), rather than something like "lie on"...).

Q is also dense in R, why not just paint the lines x=q*(pi^i) for all y q in Q (Pi isn't algebric so all are different) and a circle projection contains an open segment therefore containig an x of all types

Your post is incomprehensible -- please use LaTeX (see e.g. http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:LaTeX_on_AoPS ), fix your various typos, etc.

There is a standard technique from set theory that solves problems like this one in an almost mindless fashion. Namely, fix a 'well-ordering' over the set of all circles in the plane (of which there are continuum many). Go through these circles in order; for each circle C in turn, color 2000 uncolored points on C in rainbow fashion. This can be done because (a) for each circle C, there are only fewer-than-continuum many circles before C in the well-ordering; (b) each (nondegenerate) circle has continuum many points, so there will be enough points 'left over' on any circle C we reach in this inductive process. Many more interesting constructions are possible in this way; one of my favorites: prove that 3-dimensional space can be partitioned into circles of unit radius.

The following solution shows how we can strengthen the problem quite a bit: Let the colors be $\{1,2,3,\dots ,k\}$. Take a surjective function $f:\mathbb R \rightarrow \{1,2,3\dots k\}$. We color point $(x,y)$ with color $f(C(y))$. Where $C$ is the conway base $13$ function. Why does it work? because if we have a circle then there is an open interval $I$ so that for every $a$ in $I$ there is a point $x,y$ on the circle with $y\in I$. The Conway function satisfies that for every interval $I$ and real number $x$ there is $i\in I$ with $f(i)=x$.