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Difference between revisions of "1978 AHSME Problems/Problem 11"

(Created page with "The circle <math>x^2 + y^2 = r</math> has center <math>(0,0)</math> and radius <math>\sqrt{r}</math>. Therefore, if the line <math>x + y = r</math> is tangent to the circle <m...")
 
 
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== Problem 11 ==
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If <math>r</math> is positive and the line whose equation is <math>x + y = r</math> is tangent to the circle whose equation is <math>x^2 + y ^2 = r</math>, then <math>r</math> equals
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<math>\textbf{(A) }\frac{1}{2}\qquad
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\textbf{(B) }1\qquad
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\textbf{(C) }2\qquad
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\textbf{(D) }\sqrt{2}\qquad
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\textbf{(E) }2\sqrt{2}    </math>
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== Solution ==
 
The circle <math>x^2 + y^2 = r</math> has center <math>(0,0)</math> and radius <math>\sqrt{r}</math>. Therefore, if the line <math>x + y = r</math> is tangent to the circle <math>x^2 + y^2 = r</math>, then the distance between <math>(0,0)</math> and the line <math>x + y = r</math> is <math>\sqrt{r}</math>.
 
The circle <math>x^2 + y^2 = r</math> has center <math>(0,0)</math> and radius <math>\sqrt{r}</math>. Therefore, if the line <math>x + y = r</math> is tangent to the circle <math>x^2 + y^2 = r</math>, then the distance between <math>(0,0)</math> and the line <math>x + y = r</math> is <math>\sqrt{r}</math>.
  
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<cmath>\frac{r}{\sqrt{2}} = \sqrt{r}.</cmath>
 
<cmath>\frac{r}{\sqrt{2}} = \sqrt{r}.</cmath>
 
Then <math>r = \sqrt{r} \cdot \sqrt{2}</math>, so <math>\sqrt{r} = \sqrt{2}</math>, which means <math>r = \boxed{2}</math> or (B), <math>2</math>.
 
Then <math>r = \sqrt{r} \cdot \sqrt{2}</math>, so <math>\sqrt{r} = \sqrt{2}</math>, which means <math>r = \boxed{2}</math> or (B), <math>2</math>.
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==See Also==
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{{AHSME box|year=1978|num-b=10|num-a=12}}
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{{MAA Notice}}

Latest revision as of 11:26, 13 February 2021

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), $2$.

See Also

1978 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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