Intersection of a circle and a parabola

Find where the circle $x^2 + y^2 = 2$ meets the parabola $y = x^2$. Substituting $x^2 = y$ into the circle gives $y + y^2 = 2$, that is $y^2 + y - 2 = (y + 2)(y - 1) = 0$. Since $y = x^2 \ge 0$, the root $y = -2$ is rejected, leaving $y = 1$.

Putting $y = 1$ back into $y = x^2$ gives $x^2 = 1$, so $x = \pm 1$. The two intersection points are $(1, 1)$ and $(-1, 1)$, shown by the large dots. The circle is drawn as an upper half $y = \sqrt{2 - x^2}$ and a lower half $y = -\sqrt{2 - x^2}$.