2018 AMC 12A Problem 16

Question

Which of the following describes the set of values of

a

for which the curves

x^2 + y^2 = a^2

and

y = x^2 - a

in the real

xy

-plane intersect at exactly

3

points?

a = \tfrac14

\tfrac14 \lt a \lt \tfrac12

a \gt \tfrac14

a = \tfrac12

a \gt \tfrac12

From

y = x^2 - a

substitute

x^2 = y + a

into the circle equation to get a quadratic in

y

One intersection is the vertex tangency at

x = 0;

two more require

x^2 = 2a - 1

to have real nonzero solutions

Substituting

x^2 = y + a

into

x^2 + y^2 = a^2

gives

y^2 + y + (a - a^2) = 0,

which factors as

(y + 1 - a)(y + a) = 0,

so

y = a - 1

or

y = -a.

These correspond to

x^2 = 2a - 1

and

x^2 = 0.

The equation

x^2 = 0

always gives the single point

(0, -a),

the vertex of the parabola. The equation

x^2 = 2a - 1

gives two more points exactly when

2a - 1 \gt 0,

i.e.

a \gt \tfrac12.

So there are

3

intersection points precisely when

a \gt \tfrac12.

Thus, the correct answer is E.

Problem 16 in other years