AMC10 2015 10B, Problem 12

For how many integers $x$ is the point $(x, -x)$ inside or on the circle of radius $10$ centered at $(5, 5)$?

Solution: The equation of the circle is $(x-5)^2+(y-5)^2=100$. Plugging in the given conditions we have $(x-5)^2+(-x-5)^2 \leq 100$. Expanding gives: $x^2-10x+25+x^2+10x+25\leq 100$, which simplifies to $x^2\leq 25$ and therefore $x\leq 5$ and $x\geq -5$. So $x$ ranges from $-5$ to $5$, for a total of $\fbox{11}$ integer values. Note by Williamgolly: Alternatively, draw out the circle and see that these points must be on the line $y=-x$.

Answer: 11