2015 AMC 10B Problems/Problem 12

Problem

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

$\textbf{(A)} 11\qquad \textbf{(B)} 12\qquad \textbf{(C)} 13\qquad \textbf{(D)} 14\qquad \textbf{(E)} 15$

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 $x\leq 5$ or $x\geq -5$ . So $x$ ranges from $-5$ to $5$ , for a total of $\boxed{\textbf{(A)} 11}$ values.

2015 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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2015 AMC 10B Problems/Problem 12

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