jmc

We complete the square.

The square whose non-constant terms agree with $x^2+1300x+1300$ is $(x+650{)}^2$. Specifically, we have $$(x+650{)}^2=x^2+1300x+650^2,$$so $$\begin{array}{rl}{x}^2+1300x+1300& =(x+650{)}^2-650^2+1300\\ & =(x+650{)}^2-650\cdot 650+2\cdot 650\\ & =(x+650{)}^2+(-650+2)\cdot 650\\ & =(x+650{)}^2+(-648)(650).\end{array}$$This has the target form $(x+b{)}^2+c$, where $b=650$ and $c=(-648)(650)$. Therefore, $\frac{c}{b}=\frac{(-648)(650)}{650}=\boxed{-648}$.