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Difference between revisions of "1969 AHSME Problems/Problem 21"

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== Solution 2 ==
 
== Solution 2 ==
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Since <math>x^2+y^2=m</math> has [[radius]] <math>\sqrt{m}</math>, and <math>x+y=\sqrt{2m}</math> is a [[diagonal]] [[line]] in the [[plane]] with [[slope]] <math>-1</math>, if the two [[curves]] are [[tangent]] then <math>x+y=\sqrt{2m}</math> either has to pass through <math>(\frac{\sqrt{m}}{2}, \frac{\sqrt{m}}{2})</math> or <math>(-\frac{\sqrt{m}}{2}, -\frac{\sqrt{m}}{2})</math>. However, as <math>\sqrt{2m}</math> is [[positive]], the [[line]] can only pass through <math>(\frac{\sqrt{m}}{2}, \frac{\sqrt{m}}{2})</math>, which it does for all <math>m</math>. Thus the answer is <math>\boxed{\textbf{(E)}}</math>.
+
Since <math>x^2+y^2=m</math> has [[radius]] <math>\sqrt{m}</math>, and <math>x+y=\sqrt{2m}</math> is a [[diagonal]] [[line]] in the [[plane]] with [[slope]] <math>-1</math>, if the two [[curves]] are [[tangent]] then <math>x+y=\sqrt{2m}</math> either has to pass through <math>(\frac{\sqrt{2m}}{2}, \frac{\sqrt{2m}}{2})</math> or <math>(-\frac{\sqrt{2m}}{2}, -\frac{\sqrt{2m}}{2})</math>. However, as <math>\sqrt{2m}</math> is [[positive]], the [[line]] can only pass through <math>(\frac{\sqrt{2m}}{2}, \frac{\sqrt{2m}}{2})</math>, which it does for all <math>m</math>. Thus the answer is <math>\boxed{\textbf{(E)}}</math>.
  
 
== See Also ==
 
== See Also ==

Latest revision as of 00:37, 13 November 2023

Problem

If the graph of $x^2+y^2=m$ is tangent to that of $x+y=\sqrt{2m}$, then:

$\text{(A) m must equal } \tfrac{1}{2}\quad \text{(B) m must equal } \frac{1}{\sqrt{2}}\quad\\ \text{(C) m must equal } \sqrt{2}\quad \text{(D) m must equal } 2\quad\\ \text{(E) m may be any non-negative real number}$

Solution 1

Note that the first equation represents a circle and the second equation represents a line. If a line is tangent to a circle, then it only hits at one point, so there will only be one solution to the system of equations.

In the second equation, $y = -x + \sqrt{2m}$. Substitution results in \[x^2 + x^2 - 2x\sqrt{2m} + 2m = m\] \[2x^2 - 2x\sqrt{2m} + m = 0\] In order for the system to have one solution, the discriminant must equal $0$. \[4(2m) - 4 \cdot 2 \cdot m = 0\] \[0 = 0\] Thus, $m$ can be a non-negative real number, so the answer is $\boxed{\textbf{(E)}}$.

Solution 2

Since $x^2+y^2=m$ has radius $\sqrt{m}$, and $x+y=\sqrt{2m}$ is a diagonal line in the plane with slope $-1$, if the two curves are tangent then $x+y=\sqrt{2m}$ either has to pass through $(\frac{\sqrt{2m}}{2}, \frac{\sqrt{2m}}{2})$ or $(-\frac{\sqrt{2m}}{2}, -\frac{\sqrt{2m}}{2})$. However, as $\sqrt{2m}$ is positive, the line can only pass through $(\frac{\sqrt{2m}}{2}, \frac{\sqrt{2m}}{2})$, which it does for all $m$. Thus the answer is $\boxed{\textbf{(E)}}$.

See Also

1969 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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 • 31 • 32 • 33 • 34 • 35
All AHSME Problems and Solutions

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