Let $X$ and $Y$ be the points where the incircle touches $A B$ and $B C$, respectively. Then $A X I$ and $A Y B$ are similar right triangles. Since $I$ is the incenter, we have $I X=I Y=2$. Using the Pythagorean theorem on triangle $A X I$, we find $A X=\sqrt{5}$. By similarity, $A Y / A X=B Y / I X$. Plugging in the numbers given, $5 / \sqrt{5}=B Y / 2$, so $B Y=2 \sqrt{5} . Y$ is the midpoint of $B C$, so $B C=4 \sqrt{5}$.