1960 AMC 12 Problem 5

Attempt Problem 5 of the 1960 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1960 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

The number of distinct points common to the graphs of

x^2+y^2=9

and

y^2=9

is:

infinitely many

four

two

one

none

Substitute

y^2=9

into the circle equation

After finding

x,

count the possible values of

y

Substitution gives

x^2+9=9,

x=0.

Then

y^2=9

gives

y=3

y=-3.

Hence there are two common points.

Therefore, the correct answer is C.

Problem 5 in other years