@Blue was kind with his comments on a previous question here.
I'd now like to share some new relationships I found using algegra and my favorite formula for generating Pythagorean triples. In this example, I'm using $\,F(n,k)=F(1,1)=(3,4,5),\,\,$ but it works for all such triples. Oddly enough, the radius of the right angle circle is also the inradius of the triangle.
$\textbf{UPDATE}$
Per comment from @David K, Using Euclid's formula $\,A=p^2-q^2\quad B=2pq\quad C=p^2+q^2.$ \begin{align*} r_1&=\dfrac{A+C-B}{2}=\dfrac{(p^2-q^2)+(p^2+q^2)-(2pq)}{2} =p (p - q)\\ r_2&=\dfrac{A+B-C}{2}=\dfrac{(p^2-q^2)+(2pq)-(p^2+q^2)}{2} =(p-q)q\\ r_3&=\dfrac{B+C-A}{2}=\dfrac{(2pq)+(p^2+q^2)-(p^2-q^2)}{2} =(p + q)q \end{align*}