UTM Ideals Varieties and Algorithm - Chapter 1 Section 2 Exercise 3

Problem:

Solution:

It is the intersection of a circle with a hyperbola, so there are only four points, there is no point 'drawing' the points.

To solve the intersection points, note that $x^2 + y^2 = 4$ and $x^2y^2 = 1$.
That gives

$\begin{eqnarray*} x^2(4 - x^2) &=& 1 \\ x^4 - 4x^2 + 1 &=& 0 \\ x^2 &=& 2 \pm \sqrt{3} \\ x &=& \pm \sqrt{2 \pm \sqrt{3}} \\ \end{eqnarray*}$

We could have solve $y$ by back substituting, but here we can simply use a symmetric argument to conclude the four points must be:

$(\sqrt{2 + \sqrt{3}}, \sqrt{2 - \sqrt{3}})$
$(\sqrt{2 - \sqrt{3}}, \sqrt{2 + \sqrt{3}})$
$(-\sqrt{2 + \sqrt{3}}, -\sqrt{2 - \sqrt{3}})$
$(-\sqrt{2 - \sqrt{3}}, -\sqrt{2 + \sqrt{3}})$