1960 AHSME Problems/Problem 5

Problem

The number of distinct points common to the graphs of $x^2+y^2=9$ and $y^2=9$ is:

$\textbf{(A) }\text{infinitely many}\qquad \textbf{(B) }\text{four}\qquad \textbf{(C) }\text{two}\qquad \textbf{(D) }\text{one}\qquad \textbf{(E) }\text{none}$

Solution

Solve the second equation by taking the square root of both sides. $y^2=9$ $y=\pm3$

Solve the first equation by substituting $y^2$ into the first equation then solving for $x$ . $x^2+9=9$ $x^2=0$ $x=0$

The two solutions are $(0,3)$ and $(0,-3)$ , so the answer is $\boxed{\textbf{(C)}}$ .

1960 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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
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1960 AHSME Problems/Problem 5

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