We know that for a given in-radius there exist at least one right triangle.
Is there any way we can find the no of distinct right triangles with integer sides (distinct if any of the three sides differ) possible with side lengths for a given in-radius.
On
The number of distinct right triangles circumscribed to a given circle is infinite.
Proof: consider circle with center $(1,1)$ and radius one.
Fix $A=(0,0)$, and take any point $B(b,0)$ on $Ox$ axis with any $b>2$. Draw a tangent to the circle that intersects axis $Oy$ in $C(0,c)$. Triangle $ABC$ is a solution.
On
We begin with a formula that generates all Pythagorean triples where $\space GCD(A,B,C)\space$ is an odd square. This includes all primitives where $\,GCD(A,B,C)=1).$ \begin{align*} &A=(2n-1)^2+&&2(2n-1)k\\ &B= &&2(2n-1)k+2k^2\\ &C=(2n-1)^2+&&2(2n-1)k+2k^2 \end{align*} The inradius of any triple generated with this formula is $\quad r=(2n-1)k.\quad$
We can now see that there are as many triples with a given inradius as there are combinations of the cofactors $\space(2n-1)\space$ and $\space k.\quad$ Note: Since $\,2n-1\,$ ia always the $\,n^{th}\,$ odd number, the value of $\,r\,$ must contain an odd cofactor for there to be two-or-more solutions. Examples:
$r=1\longrightarrow\quad F(1,1)=(3,4,5)$
$r=2\longrightarrow\quad F(1,2)=(5,12,13)$
$r=3\longrightarrow\quad F(1,3)=(7,24,25),\,F(2,1)=(15,8,17)$
$r=4\longrightarrow\quad F(1,4)=(9,40,41)$
$r=5\longrightarrow\quad F(1,5)=(11,60,61),\,F(3,1)=(35,12,37)$
$r=6\longrightarrow\quad F(1,6)=(13,84,85),\,F(2,2)=(21,20,29)$
$\qquad\vdots$
$r=9\longrightarrow\quad F(1,9)= (19,180,181) ,\,F(2,3)=(27,36,45) ,\,F(5,1)=(323,36,325)$
Hint: draw a right angle, and inscribe a circle of prescribed radius within it, and try to draw a few lines tangent to the circle.