Are rational points dense on every circle in the coordinate plane?

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.

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 $\bmod 4$.

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 $3$.