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 $(5, 5)$ ?

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

Video Solution

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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 $\boxed{\mathbf{(A)}\ 11}$ integer values.

Note by Williamgolly: Alternatively, draw out the circle and see that these points must be on the line $y=-x$ .

2015 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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All AMC 10 Problems and Solutions

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2015 AMC 10B Problems/Problem 12

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