You have
$$b=\frac{2r\left(a-r\right)}{a-2r}$$
You can rewrite this as:
$$b = r - \frac{a}2 + \frac{a^2}{2(a-2r)}$$
For this to be an integer, you want $a^2$ to be divisible by $a-2r$. So if $a$ is given, look for the divisors $d|a^2$ for which $d<a$, and then let $r=\frac{a-d}2$. Of course you want this to be an integer too, so you also want $d$ to have the same parity are $a$.
Note that not all of these divisors will necessarily work, not only because these don't always give integer solution due to those halves in the expression, but also because there is no guarantee that even with integer $a$, $b$ and $r$ that they actually give a Pythagorean triangle.
This is all easy to do even for very large $a$ provided you have its prime factorization.
This also suggests that there will be several values of $r$ which could work with larger $a$, so that your conjecture is probably false even when restricted to only Pythagorean triangles.
Edit:
Note that Pythagorean triangles always have integer inradius, so your conjecture essentially says that no two Pythagorean triangles have the same shortest side. Here are two (primitive) Pythagorean triangles with the same shortest side: $(20, 21, 29)$ and $(20, 99, 101)$. Here is another pair sharing a shortest side with an odd length: $(105, 208, 233)$ and $(105,608,617)$.
Your formula does generate Pythagorean triples but misses most of them and appears to require seeds to work.
I'm not sure what you are generating. You do generate triples where $C-A=2$ in the first column but that can be generated more easily by
$\quad A=4n^2-1\quad B=4n\quad C=4n^2+1.\quad $ The rest of the table shows no pattern that I can see, like a consistent side difference within a set or consistent increment of side values within a set. The following formula generates all primitives and a few that are not but there is a consistent
$C-B=(2n-1)^2\quad$ and $\quad A_{n+1}-A_{n}=2(2n-1).\quad$ It is the formula derive when
$A=(2n-1+k)^2-k^2,\space
B=2(2n-1+k)k,\space
C=(2n-1+k)^2+k^2$
\begin{align*}
A=(2n-1)^2+ \quad &2(2n-1)k\\
B=\hspace{55pt} &2(2n-1)k\quad+2k^2\\
C=(2n-1)^2+ \quad &2(2n-1)k\quad +2k^2
\end{align*}
Here is a sample of what it generates
$$\begin{array}{c|c|c|c|c|}
n & k=1 & k=2 & k=3 & k=4 \\ \hline
Set_1 & 3,4,5 & 5,12,13& 7,24,25& 9,40,41 \\ \hline
Set_2 & 15,8,17 & 21,20,29 &27,36,45 &33,56,65 \\ \hline
Set_3 & 35,12,37 & 45,28,53 &55,48,73 &65,72,97 \\ \hline
Set_{4} &63,16,65 &77,36,85 &91,60,109 &105,88,137 \\ \hline
Set_{5} &99,20,101 &117,44,125 &135,72,153 &153,104,185 \\ \hline
\end{array}$$
Your formula generates the first column but nothing with a pattern I can see in the other cells. If you do want to work with areas, there is a list of them here. If you can figure out how to generate this sequence, I can show you how to find all of the
$1,\space 2, \text{ or } 3\space $ triples
that correspond to each area.
Best Answer
For point $A$ $\left(-\frac{15}{17},\frac{8}{17}\right)$ for example, you know from the distance formula that $$\left(-\frac{15}{17}\right)^2 + \left(\frac{8}{17}\right)^2 = 1^2$$ so $$\left(-{15}\right)^2 + \left({8}\right)^2 = 17^2$$ or equally $$8^2 + 15^2 = 17^2$$ and so $(8,15,17)$ is a Pythagorean triple. A similar process of multiplying up the denominators produces triples for the other given points
Now the reverse process of identifying points on the circle given a Pythagorean triple should be clear. $(3,4,5)$ can for example produce a point like $\left(\frac{4}{5},\frac{3}{5}\right)$.