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 $.

A system of equations such as yours, namely
\begin{cases}
x^2+y^2=4 \\
xy=1
\end{cases}
is symmetric, because it doesn't change when $x$ and $y$ are swapped.
Rewrite $x^2+y^2=4$ as $(x+y)^2-2xy=4$, so you get
\begin{cases}
(x+y)^2=6 \\
xy=1
\end{cases}
that can be divided into
$$
\begin{cases}
x+y=\sqrt{6} \\
xy=1
\end{cases}
\qquad\text{or}\qquad
\begin{cases}
x+y=-\sqrt{6} \\
xy=1
\end{cases}
$$
Solving the former suffices, for any solution of the former system provides a solution of the latter by changing signs.
The problem is now to find two numbers we know the sum and product of, that is, the roots of the quadratic equation
$$
z^2-\sqrt{6}\,z+1=0
$$
The roots are
$$
z=\frac{\sqrt{6}-\sqrt{2}}{2}
\qquad\text{or}\qquad
z=\frac{\sqrt{6}+\sqrt{2}}{2}
$$
Thus we get the four points
$$
\left(\frac{\sqrt{6}-\sqrt{2}}{2},\frac{\sqrt{6}+\sqrt{2}}{2}\right)
\quad
\left(\frac{\sqrt{6}+\sqrt{2}}{2},\frac{\sqrt{6}-\sqrt{2}}{2}\right)
\\
\left(-\frac{\sqrt{6}-\sqrt{2}}{2},-\frac{\sqrt{6}+\sqrt{2}}{2}\right)
\quad
\left(-\frac{\sqrt{6}+\sqrt{2}}{2},-\frac{\sqrt{6}-\sqrt{2}}{2}\right)
$$
