I've been working on it for a few days, and I can't seem to find a possible answer. My maths teacher told me it was not impossible (and she's not the joking kind).

$\displaystyle x^2 + y^2 = 4$

$\displaystyle y= -x+3$

So let me show you what I tried...

$\displaystyle x^2 + y^2 = 4$

$\displaystyle x^2 + (-x+3)^2 = 4$

$\displaystyle x^2 + (-x+3)(-x+3) = 4$

$\displaystyle x^2 + (x^2-3x-3x+9) = 4$

$\displaystyle 2x^2 -6x + 5 = 0$

Then...

$\displaystyle \frac {-b \pm \sqrt {b^2-4ac}}{2a}$ so $\displaystyle \frac {6 \pm \sqrt {36-40}}{4}$ so $\displaystyle \frac {6 \pm \sqrt {-4}}{4}$

And $\displaystyle \sqrt {-4}$ is obviously the problem here.

Any ideas? Thanks.