2018 AMC 10A Problem 21

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 = \dfrac14

\dfrac14 \lt a \lt \dfrac12

a \gt \dfrac14

a = \dfrac12

a \gt \dfrac12

Substitute the parabola equation into the circle equation

The nonzero roots appear only when

2a-1

is positive

Substitute

y=x^2-a

into

x^2+y^2=a^2

. This gives

x^2+(x^2-a)^2=a^2

, so

x^2(x^2-(2a-1))=0

.

The factor

x^2=0

always gives the single point

(0,-a)

. The other factor gives two additional real points exactly when

2a-1>0

.

There are exactly three intersection points when

a>\dfrac12

. Thus, E is the correct answer.

Problem 21 in other years