Q:

The intersection of a line with a circle

Get the intersections of the line $y=x+2$ with the circle $x^2+y^2=10$
What I did:
$y^2=10-x^2$ 
$y=\sqrt{10-x^2}$ or $y=-\sqrt{10-x^2}$
$ x+ 2 = y=\sqrt{10-x^2}$  
If you continue, $x=-3$ or $x=1$ , so you get 2 points $(1,3)$, $(-3,-1)$
But then, and here is where the problems come:
$x+2=-\sqrt{10-x^2}$
I then, after a while get the point $(-3\dfrac{1}{2}, -1\dfrac{1}{2}$) but this doesn't seem to be correct. What have I done wrong at the end?

A:

If you do everything correct, the solution of $x+2=-\sqrt{10-x^2}$ would be almost the same - $(1,-3)$, $(-3,1)$. It's refer to intersaction of $y = -(x+2)$ and original circle.
By the way, your answer is correct.
I have no idea how you'd get the point $(-3\dfrac{1}{2}, -1\dfrac{1}{2})$.

