I will use the approach of Pythagorean triples $a^2+b^2=c^2=2009^2$. In any _primitive_ triple ($a,b,c$ are coprime), all prime factors of $c$ are of the form $4k+1$, but $2009=7×7×41$. Thus the values of $a$ and $b$ must themselves be multiples of 49, and we get $$(49c)^2+(49d)^2=(49\cdot41)^2$$ $$c^2+d^2=41^2$$ where $c$ and $d$ are coprime. Now there are only two solutions for $(c,d)$, $(9,40)$ and $(40,9)$. Therefore the number of tangents with integer lengths is also two, and these are of lengths $49\cdot9=441$ and $49\cdot40=1960$.