# 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 $x$.

• Just curious: what do you mean by the graphical method? How do you read out the value of the coordinates of the intersection points? Nov 22, 2016 at 12:24
• By drawing respective graphs and then finding the points of intersections. Nov 27, 2016 at 17:06

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

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

• we will then get 4 points right? Nov 21, 2016 at 17:11
• @Batwayne Correct $x = \{2(2 + 3^{1/2})^{1/2}, 2(2 - 3^{1/2})^{1/2}, -2(2 + 3^{1/2})^{1/2}, -2(2 - 3^{1/2})^{1/2} \}$ and $y = 1/x$ Nov 21, 2016 at 17:12

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.

• nice!$\quad\qquad$ Nov 22, 2016 at 5:30

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

• You could simplify the $y$ coordinates using $1/(2\pm\sqrt{3}) = 2\mp\sqrt{3}$, which also shows the symmetry more clearly. Nov 21, 2016 at 21:22
• I could. But I personally don't like that. It treats the pm sign as a distinct variable rather than a parity sign. Just my opinion. Anyway, that's why I listed the four points, distinctly. Nov 21, 2016 at 21:32
• I actually meant writing the points as $(\sqrt{2+\sqrt{3}},\sqrt{2-\sqrt{3}})$ etc Nov 21, 2016 at 21:36
• Oh, you're right. I maybe should have. Nov 21, 2016 at 21:39

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 $$t=\pm\frac12\text{arcosh }2=\pm\frac12\ln(2+\sqrt3)=\pm\ln\sqrt{2+\sqrt3}.$$

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

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

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 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 $$\left(\frac{\sqrt{6}-\sqrt{2}}{2},\frac{\sqrt{6}+\sqrt{2}}{2}\right) \quad \left(\frac{\sqrt{6}+\sqrt{2}}{2},\frac{\sqrt{6}-\sqrt{2}}{2}\right) \\ \left(-\frac{\sqrt{6}-\sqrt{2}}{2},-\frac{\sqrt{6}+\sqrt{2}}{2}\right) \quad \left(-\frac{\sqrt{6}+\sqrt{2}}{2},-\frac{\sqrt{6}-\sqrt{2}}{2}\right)$$