Number of height-limited rational points on a circle Consider origin-centered circles $C(r)$ of radius $r \le 1$.
I am seeking to learn how many rational points might lie on $C(r)$,
where each rational point coordinate has height $\le h$.
For example, these are the rationals in $[0,1]$ with $h \le 5$:
$$
\left(
0,\frac{1}{5},\frac{1}{4},\frac{1}{3},\frac{2}{5},\frac{1}{2},\frac{3}{5},
\frac{2}{3},\frac{3}{4},\frac{4}{5},1
\right)
$$
Rational points of height $\le h$ have both coordinates from this list,
multiplied by $\pm 1$.

Q. What is the growth rate of the maximum number of rational points
  of height $\le h$ on $C(r)$, $r \le 1$, as a function of $h$?

Here is a bit of data up to $h=20$:

          


For example, for $h=7$, $C(\frac{5}{7})$ 
passes through
these $12$ points:
$$
\left(
 -\tfrac{4}{7} , -\tfrac{3}{7} 
\right),
\left(
 -\tfrac{3}{7} , -\tfrac{4}{7}
\right),
\left(
 0 , -\tfrac{5}{7}
\right),
\left(
 \tfrac{3}{7} , -\tfrac{4}{7}
\right),
\left(
 \tfrac{4}{7} , -\tfrac{3}{7} 
\right),
\left(
 \tfrac{5}{7} , 0 
\right),
$$
$$
\left(
 \tfrac{4}{7} , \tfrac{3}{7} 
\right),
\left(
 \tfrac{3}{7} , \tfrac{4}{7} 
\right),
\left(
 0 , \tfrac{5}{7} 
\right),
\left(
 -\tfrac{3}{7} , \tfrac{4}{7} 
\right),
\left(
 -\tfrac{4}{7} , \tfrac{3}{7} 
\right),
\left(
 -\tfrac{5}{7} , 0 
\right)
$$
If I've calculated correctly, no circle passes through more than $12$
points of height $\le 7$.
Circles that achieve these maxima are illustrated below.

          


          

Background points are the rational points of height $h \le 20$.


Added. Since the radii that achieve the maxima I found
for $13 \le h \le 20$ are all exactly $1$,
it may be that the question can be reduced to counting the
number of rational points of height $\le h$ on just specifically $C(1)$.

Answered. Lucia's answer matches even the small-$h$ data I gathered:

          



I'll content myself with counting the number of points on $C(1)$ (which should surely be close to the maximum) -- the answer is quite nice, it is about $ \frac{4}{\pi } h$.  
To see this, note that we are counting essentially Pythagorean triples $u^2-v^2, 2uv, u^2+v^2$, with $u^2+v^2\le h$ and we may suppose that $u$ and $v$ are non-negative, that $u$ and $v$ are coprime, and that $u^2+v^2$ is odd.  The lattice point count we need is four times this number, since we must also count the lattice point $(2uv/(u^2+v^2),(u^2-v^2)/(u^2+v^2))$ (in addition to $((u^2-v^2)/(u^2+v^2),2uv/(u^2+v^2))$, and we must also allow the $2uv/(u^2+v^2)$ coordinate to be negative).  
Thus to summarize we want 
$$
4 \sum_{n\le h, n \text{ odd }} R(n), 
$$ 
where $R(n)$ is the number of ways of writing $n$ as $u^2+v^2$ with both $u$ and $v$ non-negative and coprime (taking care to set $R(1)$ to be $1$).
  A little number theory, going back to Fermat, gives that $R(n)$ is a multiplicative function with $R(2^k)=0$ 
(so we don't have to worry about $n$ odd anymore), $R(p^k)=2$ for $p\equiv 1\pmod 4$ and $k\ge 1$, and $R(p^k)=0$ if $p\equiv 3\pmod 4$.  For example if $h=20$, then $R(1)=1$, $R(5)=2$, $R(13)=2$, and $R(17)=2$ and the rest are zero, and the number here is $28$ as in the numerics. 
From here a standard argument (or one can do this via counting lattice points in a circle) leads to the asymptotic 
$$ 
4 \sum_{n\le h} R(n) \sim 4 \frac{1}{2} \prod_{p\equiv 1 \pmod 4} \Big(1+\frac{2}{p}+\frac{2}{p^2}+\ldots \Big) \Big(1-\frac 1p\Big) 
\prod_{p\equiv 3\pmod 4} \Big(1-\frac 1p\Big) h, 
$$
and the above simplifies (using $1-1/3+1/5-1/7+\ldots =\pi/4$ and $1/1^2+1/3^2+1/5^2+\ldots = \pi^2/8$) to give 
$$ 
\sim 2 \frac{\pi/4}{\pi^2/8} h = \frac{4}{\pi} h. 
$$
One should be able to refine this to count lattice points on other circles as well, and thus show that radius $1$ does achieve the maximum.
