jmc

Add $-r^2-2r-4$ to both sides of the given congruence to obtain $2r\equiv -3(\mathrm{mod}55)$. We can multiply both sides by $28$ to get $56r\equiv -3\cdot 28(\mathrm{mod}55)$. Subtracting $55r$ from the left-hand side and adding $2\cdot 55=110$ to the right-hand side gives $r\equiv 26(\mathrm{mod}55)$. Thus $r=26+55k$ for some integer $k$. Solving $26+55k\ge 1000$, we find that $k=18$ is the smallest value of $k$ for which $r$ has four digits. Thus the minimum four-digit value of $r$ is $26+55(18)=\boxed{1016}$.