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: 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.
A: 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: $(\sqrt{2 + \sqrt{3}},{\sqrt{2 - \sqrt{3}}});(\sqrt{2 - \sqrt{3}},{\sqrt{2 +\sqrt{3}}});(-\sqrt{2 + \sqrt{3}},-{\sqrt{2 - \sqrt{3}}});(-\sqrt{2 - \sqrt{3}},-{\sqrt{2 + \sqrt{3}}});$
A: By trigonometry:
Any point on the circle has coordinates $(2\cos t,2\sin t)$. Then plugging in the other equation
$$4\sin t\cos t=1,$$
$$\sin 2t=\frac12,$$
giving
$$t\in{\frac\pi{12},\frac{5\pi}{12},\frac{13\pi}{12},\frac{17\pi}{12}}.$$

By hyperbolic trigonometry:
Let $x=e^t,y=e^{-t}$ be a parameteric solution of the equation of the hyperbola. (There is another branch with opposite signs.)
Then by the equation of the cirle
$$x^2+y^2=e^{2t}+e^{-2t}=2\cosh2t=4$$ and $$t=\pm\frac12\text{arcosh }2=\pm\frac12\ln(2+\sqrt3)=\pm\ln\sqrt{2+\sqrt3}.$$
A: 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
$$
x = \pm(2 \pm 3^{1/2})^{1/2} \quad y = 1/x
$$
A: Solve 
x^2+y^2=4
xy=1
we get 4 solutions for x,y in 1st and 3rd quadrant and symetric around 0,0
(sqrt(6)+sqrt(2))/2,(sqrt(6)-sqrt(2))/2
(sqrt(6)-sqrt(2))/2,(sqrt(6)+sqrt(2))/2
-(sqrt(6)+sqrt(2))/2,-(sqrt(6)-sqrt(2))/2
-(sqrt(6)-sqrt(2))/2,-(sqrt(6)+sqrt(2))/2
A: The circle is ,
                         $x^2 + y^2 = 4\tag{1}$
The hyperbola is,
$xy=1\tag{2}$
Adding 2xy on both sides of equation 1,  
$$x^2+y^2+2xy=4+2xy$$  $$(x+y)^2=4+2xy$$
$$(x+y)^2=6\tag{since xy=1,from (1)}$$
$$x+y=\pm\sqrt6\tag{3}$$
putting 2 into 3 ,
$$x+1/x=\pm\sqrt6$$
$$x^2\pm\sqrt6x+1=0$$
solving this quadratic first for +$\sqrt6$ and then for -$\sqrt6$ we get 4 solutions for x,
$$\frac{\sqrt6+\sqrt2}2 ,\frac{\sqrt6-\sqrt2}2$$  and  $$\frac{-\sqrt6+\sqrt2}2,\frac{-\sqrt6-\sqrt2}2$$
the solutions for y will be inverse of each of the solutions for x
A: By a change of variable, $$x^2+y^2=4,\\xy=1$$ can be rewritten
$$X+Y=4,\\XY=1$$ provided you keep in mind that $x$ and $y$ have the same sign.
Then this is a classical sum/product problem, solved by
$$(X-Y)^2=(X+Y)^2-4XY=12,$$ then
$$X,Y=\frac{4\pm\sqrt{12}}2=2\pm\sqrt3.$$
Finally there are four solutions in $x,y$,
$$x=\color{blue}\pm\sqrt{2\color{green}\pm\sqrt3},\\y=\color{blue}\pm\sqrt{2\color{green}\mp\sqrt3}$$

More directly, multiply by $x^2$ and
$$x^2+y^2=4\implies x^4+x^2y^2=x^4+1=4x^2,$$ which is a biquadratic equation.

If you want, you can unnest the radicals by
$$\sqrt{2\pm\sqrt3}=\sqrt{\frac{4\pm2\sqrt3}2}=\sqrt{\frac{(\sqrt3\pm1)^2}2}=\frac{\sqrt3\pm1}{\sqrt2}.$$
A: 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
$$
\left(\frac{\sqrt{6}-\sqrt{2}}{2},\frac{\sqrt{6}+\sqrt{2}}{2}\right)
\quad
\left(\frac{\sqrt{6}+\sqrt{2}}{2},\frac{\sqrt{6}-\sqrt{2}}{2}\right)
\\
\left(-\frac{\sqrt{6}-\sqrt{2}}{2},-\frac{\sqrt{6}+\sqrt{2}}{2}\right)
\quad
\left(-\frac{\sqrt{6}+\sqrt{2}}{2},-\frac{\sqrt{6}-\sqrt{2}}{2}\right)
$$
