2018 AMC 10A Problem 21

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.