Are rational points dense on every circle in the coordinate plane? Are rational points dense on every circle in the coordinate plane?
First thing first I know that rational points are dense on the unit circle.
However, I am not so sure how to show that rational points are not dense on every circle.
How would one come about answering this.
Any hits are appreciate it.
 A: When will there be a dense subset of rational points in a circle?
If $x^2+y^2=r^2$ has any rational point, then the rational points in it are dense in it. 
More generally, the following are equivalent about a (non-degenerate) circle in $\mathbb R^2:$


*

*The set of rational points on a circle are dense in the circle

*The circle has a rational center and a rational point

*The circle has three rational points.


I'll outline why $2\implies 1.$ We can assume that the center of your circle is $(0,0).$ Your circle has an equation like:
$$x^2+y^2=r^2$$
Since it has a rational point, it also means $r^2$ is rational.
Now, if $(x_1,y_1)$ is your rational point, take any line through that point with a rational slope, $m.$ Then the set of pairs $(x_1+t,y_1+mt)$ will (except when $m$ is the tangent of the circle at $(x_1,y_1)$) hit the circle again. But that yields a rational quadratic equation fo $t$ with a known rational root, $t=0.$ So the other root $t$ is also rational, and the other point is rational.
$3\implies 2$ is because finding the circumcenter of three points is a linear process.
And $1\implies 3$ because $(1)$ means there are infinitely many rational points on the circle, so at least $3.$
A: In particular, $x^2+y^2=3$ cannot have any rational points.  If it had any such points then there would be integers $a,b,c$ having no common factor such that $(a/c)^2+(b/c)^2=3$ therefore $a^2+b^2=3c^2$.  But with no common factor at least one of $a,b,c$ must be odd and all possibilities conforming with this requirement fail $\bmod 4$.
A: For a little more detail to Oscar's answer, the reason we may require that $a,$ $b$, and $c$ are co-prime is that if $$\left(\frac{a}{b}\right)^2 + \left(\frac{c}{d}\right)^2 = 3,$$ we may write $$(ad)^2 + (bc)^2 = 3 (bd)^2.$$  Hence $(ad)^2 = b^2(3 d^2 - c^2),$ so $b^2$ divides $(ad)^2$ and $b$ divides $ad$.  
With this, we may write $ad = bk$ and divide the $b^2$ from both sides, getting $$k^2 + c^2 = 3d^2,$$ at which point we may apply Oscar's $\mod 4$ argument.
Also, this example is necessary to flesh out Matt's answer: he ends with

Then $r^2$ must be a sum of two squares, which is not true of all integers.

However, in his setup, we require that $r$ satisfies "$r^2 \in \mathbb{N}$ but $r^2 k^2$ is not a sum of two squares for all $k \in \mathbb{N}$," and it's not clear that such an $r$ exists until you establish an example like $3$.
A: They're not. No two different circles centered at the origin contain any of the same points. There are uncountably many circles (specifically, one for each real number, corresponding to the radius) , so most circles contain no rational points at all.
We can find some more specific examples. Specifically, any rational point $(a, b)$ on a circle of radius $r$ centered at the origin satisfies $a^2+b^2=r^2$. In particular, $r^2$ must be rational. There are also radii whose squares are rational where there are no rational points. Clearing denominators, say multiplying by some $c^2$ to do so, we have that $c^2r^2$ is a sum of two squares. If $r^2$ is an integer, then $r^2$ must be a sum of two squares, since an integer is a sum of two squares if and only if its prime factorization doesn't contain an odd power of a prime congruent to $3$ mod $4$. $r^2$ was arbitrary, so if we choose it not to be a sum of two squares we get circles with no rational points.
A: First of all, I want to thank @Hidaw for having answered such an interesting question, which was swirling around in my head some days ago. After reading all the answers to this post with great pleasure, I discovered that Humke and Krajewski completely solved the problem in a very simple and beautiful article on the American Mathematical Monthly in 1979 (a few years before I was born!) called A Characterization of Circles Which Contain Rational Points.
In particular they establish the very relevant result that if the radius $r$ is irrational, but $r^2=p/q$, with $p, q \in \mathbb{Q}$, then the circle $C$ of equation $x^2+y^2=r^2$ contains no rational point if $pq$ is not the sum of two square integers, while $\mathbb{Q}$ is dense on $C$ if $pq$ is the sum of two square integers.
