The value(s) of y for which the following pair of equations x^2+y^2-16&=0, x^2-3y+12&=0 may have a real common solution, are:

1972 AMC 12 Problem 11

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11.

The value(s) of

y

for which the following pair of equations

\begin{aligned} x^2+y^2-16&=0,\\ x^2-3y+12&=0 \end{aligned}

may have a real common solution, are:

4

only

-7,

4

0,

4

y

all

y

Eliminate

x^2

between the equations

After finding the two candidate

y

-values, check whether the corresponding

x^2

is nonnegative

From the second equation,

x^2=3y-12.

Substitution into the first gives

\begin{aligned} y^2+3y-28&=(y-4)(y+7)\\ &=0. \end{aligned}

y=4,

then

x=0.

y=-7,

then

x^2=-33,

so no real

x

exists. Thus only

y=4

is possible.

Therefore, the correct answer is A.