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In right triangle $ABC$, the area is twice the perimeter, and all sides have integer lengths. Compute the sum of all possible circumradii of $ABC$.

I only have set up an equation
$$frac{xy}{2}=2left(x+y+sqrt{x^2+y^2}right)$$ and
$$R=frac{xy sqrt{x^2+y^2}}{2xy}=frac{sqrt{x^2+y^2}}{2}$$

We can parameterize the sides of a right triangle ABC right-angled at C with integer sides in the following manner:
$$a=k(x^2-y^2),
b=2kxy,
c=k(x^2+y^2)$$
where k is any positive integer, x and y are co-prime integers with $xnotequiv y (textrm{mod} 2)$.

Using the condition, $frac{ab}{2}=2(a+b+c)$, we obtain $ky(x-y)=4$ whence $k=4, y=1, x=2$ or $k=1,y=4,x=5$ or $k=2, y=2, x=3$. This leads to the triangles $(12,16,20)$, $(10,24,26)$ and $(9,40,41)$. In a right triangle, circumradius is half the hypotenuse. Therefore, $$ R=10,13, text{or} 20.5 $$

$$frac{xy}{2} = 2left(x+y +sqrt{x^2+y^2}right) implies (xy - 4(x+y))^2 = 16(x^2+y^2)\ iff xy(xy-8(x+y)+32) = 0 implies (x-8)(y-8) = 32$$
Since $x$ and $y$ are integers and $32 = 2^5$, the total numbers of factors are $(5+1)=6$ which are $1,2,4,8,16,32$ .
So possible cases avoiding any redundancy for we just want circumradii are as follows:
$$underset{32}{(x-8)}underset{1}{(y-8)} = 32$$
$$underset{16}{(x-8)}underset{2}{(y-8)} = 32$$
$$underset{8}{(x-8)}underset{4}{(y-8)} = 32$$
So possible triplets turn out to be $40,9,41$;
$24,10,26$;
$16,12,20$.
So, possible circumradii are $frac{41}2,frac{26}2,frac{20}2$.