2015 AMC 10B Problem 12

A point is inside or on the circle when its distance to the center $(5, 5)$ is at most $10$. Distances involve square roots, so compare the squares instead: the squared distance must be at most $10^2 = 100$.

1. The point $(x, -x)$ is $x - 5$ across from the center and $-x - 5$ up or down from it. By the Pythagorean theorem the condition is
$$(x - 5)^2 + (-x - 5)^2 \le 100.$$
2. Expand each square. $(x-5)^2 = x^2 - 10x + 25$. And $(-x-5)^2$ is the same as $(x+5)^2 = x^2 + 10x + 25$.
3. Add them. The $-10x$ and $+10x$ cancel:
$$2x^2 + 50 \le 100, \quad\text{so}\quad 2x^2 \le 50, \quad\text{so}\quad x^2 \le 25.$$
4. The whole numbers whose square is at most $25$ are $-5, -4, \ldots, 4, 5$. That is $5$ negatives, $5$ positives and $0$, which is $11$ numbers.

The answer is $\boxed{\textbf{(A)}\ 11}$.

"Inside or on a circle" is one inequality: squared distance to the center is at most the squared radius. When the points have a pattern like $(x, -x)$, put the pattern into that inequality and you are left with a problem in one letter. Here the $x$ terms cancel because the line $y = -x$ runs at right angles to the line from the origin to the center, so the answer is balanced around $x = 0$.

Draw it. The line $y = -x$ passes through the origin, and the closest point on that line to the center $(5, 5)$ is the origin itself, at distance $\sqrt{5^2 + 5^2} = \sqrt{50}$. The circle crosses the line where the distance from the center is $10$. By the Pythagorean theorem, those crossing points are $\sqrt{100 - 50} = \sqrt{50}$ from the origin along the line. The point $(5, -5)$ is exactly $\sqrt{50}$ from the origin, so the crossings are $(5, -5)$ and $(-5, 5)$. The integers $x$ from $-5$ to $5$ give $11$ points.