2015 AMC 10B Problem 12

The squared distance from $(x,-x)$ to $(5,5)$ is $$\begin{aligned} &(x-5)^2+(-x-5)^2 \\ &= 2x^2+50. \end{aligned}$$ Being inside or on the circle means $2x^2+50\le100$, so $x^2\le25$. Thus $-5\le x\le5$, giving $11$ integer values. Thus, the correct answer is A.