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.
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.
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.