2015 AMC 10B Problem 12

A point is inside or on the circle exactly when its distance to the center $(5,5)$ is at most $10$. Squaring, the condition for $(x, -x)$ is
$$(x-5)^2 + (-x-5)^2 \le 100.$$
Expand: $(x-5)^2 = x^2 - 10x + 25$ and $(-x-5)^2 = (x+5)^2 = x^2 + 10x + 25$. The $x$ terms cancel, leaving
$$2x^2 + 50 \le 100 \quad\Longrightarrow\quad x^2 \le 25 \quad\Longrightarrow\quad -5 \le x \le 5.$$
The integers in this range are $-5, -4, \dots, 5$, which is $11$ values.

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

"Inside or on a circle" is a single inequality in the coordinates. The points $(x, -x)$ lie on the line $y = -x$, which is perpendicular to the line from the origin through the center $(5,5)$; that symmetry is why the linear terms cancel and the answer is symmetric about $x = 0$. When a set of points is parametrized by one variable, substitute and reduce the geometry to a one-variable inequality.

Geometrically, the line $y = -x$ passes at distance $\frac{|5+5|}{\sqrt2} = 5\sqrt2$ from the center. A chord at that distance in a circle of radius $10$ has half-length $\sqrt{100 - 50} = 5\sqrt2$, so it spans from $x = -5$ to $x = 5$ on the line $y = -x$ (each unit of $x$ is $\sqrt2$ of length). Same $11$ integers.

• Counting the integers from -5 to 5 as 10 (forgetting zero or one endpoint), or using strict inequality and dropping x = 5 and x = -5.
• Writing the second coordinate as $(x-5)^2$ twice, i.e. plugging in $(x, x)$ instead of $(x, -x)$.