1978 AHSME Problems/Problem 11

Problem 11

If $r$ is positive and the line whose equation is $x + y = r$ is tangent to the circle whose equation is $x^2 + y ^2 = r$ , then $r$ equals

$\textbf{(A) }\frac{1}{2}\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }\sqrt{2}\qquad \textbf{(E) }2\sqrt{2}$

Solution

The circle $x^2 + y^2 = r$ has center $(0,0)$ and radius $\sqrt{r}$ . Therefore, if the line $x + y = r$ is tangent to the circle $x^2 + y^2 = r$ , then the distance between $(0,0)$ and the line $x + y = r$ is $\sqrt{r}$ .

The distance between $(0,0)$ and the line $x + y = r$ is $\frac{|0 + 0 - r|}{\sqrt{1^2 + 1^2}} = \frac{r}{\sqrt{2}}.$ Hence, $\frac{r}{\sqrt{2}} = \sqrt{r}.$ Then $r = \sqrt{r} \cdot \sqrt{2}$ , so $\sqrt{r} = \sqrt{2}$ , which means $r = \boxed{2}$ or (B), $2$ .

1978 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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1978 AHSME Problems/Problem 11

Problem 11

Solution

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