Can this be solved without resorting to graphical method?

The circle is described by

$$ x^2 + y^2 = 4 \tag{a} $$

and the hyperbola by

$$ y = 1/x \tag{b} $$

Replacing (b) into (a) you get

$$ x^2 + \frac{1}{x^2} = 4 \quad\Rightarrow\quad x^4 - 4x^2 + 1 = 0 $$

this is a quadratic equation in $x^2$ whose solutions are

$$ x^2 = 2 \pm \sqrt{3} $$

The intersection are then

$$ x = \pm(2 \pm 3^{1/2})^{1/2} \quad y = 1/x $$

You can also directly combine the equations into complete binomial formulas $$ (x+y)^2=x^2+y^2+2xy=4+2=6,\\ (x-y)^2=x^2+y^2-2xy=4-2=2 $$ and solve the trivial linear system for each of the 4 sign combinations of the roots.

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