1960 AHSME Problems/Problem 5

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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 $\boxed{\textbf{(C)}}$.

See Also

1960 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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