Intersection of a circle and a parabola

We find where the circle $x^2 + y^2 = 2$ meets the parabola $y = x^2$.

Eliminating the easy variable

When solving a system, eliminate whichever variable goes most easily. The parabola $y = x^2$ can be substituted straight into the circle, replacing $x^2$ by $y$ and leaving an equation in $y$ alone.

so $y = -2$ or $y = 1$.

Discarding the meaningless root

Here care is needed. Since $y = x^2$ is a square, it can never be $-2$: that root appeared along the way but carries no meaning. A solution of the algebra is not automatically a point of the figure, so always return to the original conditions and check.

Putting $y = 1$ back gives $x = \pm 1$. The intersection points are $(1, 1)$ and $(-1, 1)$; substituting into the circle confirms $1 + 1 = 2$. Both curves are symmetric about the $y$-axis, so the intersections appear as a mirror pair.

Eliminating the other variable

Substituting $y = x^2$ into the circle instead gives a quartic.

Since $x^2 = -2$ has no real solution we are left with $x = \pm 1$. The same answer, but at a higher degree and more work. Which variable you choose to eliminate often decides how heavy the algebra becomes.

Changing the radius

Repeating the computation for radius $r$ gives $y^2 + y - r^2 = 0$, which has exactly one positive root $y$, and that single $y$ yields the two points $x = \pm\sqrt{y}$. So there are always exactly two intersections, as long as the circle is centered at the origin.

Intuitively, the parabola starts at the origin, inside the circle, and runs off to infinity, so it must cross out of the circle exactly twice.

Moving the center

Shift the center off the origin along the $y$-axis and the number of intersections can rise to four: the intersections of two conics are, in general, the roots of a quartic equation¹.

The large dots on the graph are the intersections, and the circle is drawn as an upper half and a lower half.

1. Conic section, Wikipedia