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?

Let the intersection be $(a,b)$, so it must satisfy both the given eqaution.
So, $a=b+2$ also  $a^2+b^2=10$
Putting $b=a+2$ in the given circle $a^2+(a+2)^2=10$
$2a^2+4a+4=10\implies a=1$ or $-3$
If $a=1,b=a+2=3$
If $a=-3,b=-3+2=-1$
So, the intersections are $(-3,-1)$ and $(1,3)$
