Finding the number of possible right angled triangles If the inradius=$2013$ of a right angled triangle with integer sides. Find the no. of possible right angled triangles that can be formed using the above information.
I have tried $r(a+b+c)=ab$ and $a^2+b^2=c^2$ , but couldn't reach further.
Thanks in anticipation
One thing we can get is 

$CF=CD=b-r$ and $BE=BD=c-r$. So, $BC=a=b+c-2r\quad (1)$. 

By Euclide's formula we can look for a solution such that $\gcd(a,b,c)=1$ and $b=m^2-n^2$, $c=2mn$ and $a=m^2+n^2$ with, $\gcd(m,n)=1$ and not both odd. 

We can then back to $(1)$ and get 

$$a+2\cdot2013=b+c\\
m^2+n^2+2\cdot2013=m^2-n^2+2mn\\
n^2+2013=mn\to n(m-n)=2013$$  

so, $n|2013$ and once $2013=3\cdot11\cdot61$ then we have the possibilities:

$$n\in\{3,11,61,3\cdot11,3\cdot61,11\cdot61,3\cdot11\cdot61\}$$

So you just have to try all possibilities for $n$ and find a suitable $m$.

For example, for $n=3$ you will find $m=674$ and then you have one solution which is $(a,b,c)=(674^2+3^2,674^2-3^2,2\cdot3\cdot674)$. 

Now just find the others.
Answer to the question is easy: 8 primitive Pythagorean triples are there for the inradius 2013. There are no other triples (non-primitive Pythagorean triples).