<s>[INST] Can this be solved without resorting to graphical method? I need to find the points of intersection of a circle with radius $2$ and centre at $(0,0)$ and a rectangular hyperbola with equation $xy=1$. As per the topic statement is there any way to solve this without the graphical method. I have tried setting the $y$ values equal but I cant solve the resulting equation for $x $.
 [/INST] The equation of the circle is $x^2 + y^2 = 4$ and the equation of the hyperbola is $xy=1$
So the point of intersection would be a common solution to 
$xy =1$
$x^2 + y^2 = 4$
so
$y = 1/x$
$x^2 + \frac 1{x^2} = 4$
$x^4  +1 = 4x^2$
$x^4 - 4x^2 + 1 = 0$
$x^2 = \frac {4 \pm \sqrt {12}}2$ 
$x^2 = 2 \pm \sqrt 3$
$x = \pm \sqrt{2 \pm \sqrt{3}}$
$y = 1/x = \pm \frac 1{\sqrt{2 \pm \sqrt{3}}}$
$= \pm \frac 1{\sqrt{2 \pm \sqrt{3}}}\frac {\sqrt {2\mp \sqrt {3}}}{\sqrt {2\mp\sqrt{3}}} $
$=\pm \frac{\sqrt {2\mp \sqrt {3}}}{\sqrt {4-3}}=\pm {\sqrt {2\mp \sqrt {3}}}$
So there are four points: $(\sqrt{2 + \sqrt{3}},{\sqrt{2 - \sqrt{3}}});(\sqrt{2 - \sqrt{3}},{\sqrt{2 +\sqrt{3}}});(-\sqrt{2 + \sqrt{3}},-{\sqrt{2 - \sqrt{3}}});(-\sqrt{2 - \sqrt{3}},-{\sqrt{2 + \sqrt{3}}});$
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