Intersection of rectangular and polar equations Find the points of intersection for the following relations:
$x^2+y^2-20 = 0$ and $\theta = -3\pi / 4$
So the first one is obviously a circle, and the second one, converting into rectangular form, is y/x = 1, and thus y=x. Substituting y as x in the first equation, I then get $x^2 = 10$, which yields $+- \sqrt {10}$, and thus the points of intersection should then be $(\sqrt{10},\sqrt{10})$ and $(-\sqrt{10},-\sqrt{10})$. However, the answer says that it is $(\sqrt{10},-\sqrt{10})$ and $(-\sqrt{10},\sqrt{10})$. What did I do wrong?
Also, for another problem, I need to find the intersection between these two:

$4x^2+16y^2-64=0$ and $x^2+y^2=9$

Following a similar approach of substitution, I was able to get 4 points: $(20/3, 7/3), (-20/3, 7/3), (-20/3,-7/3), (20/3, -7/3)$. But the answer says $(3,0) and (-3,0)$... 

How about $x=\sqrt{20}\cos \phi$, and $y=\sqrt{20}\sin \phi$. Putting $\phi = -3 \frac{\pi}{4}$ and $\phi = -3 \frac{\pi}{4}+\pi$ gives you the answer that you mentioned $(\sqrt{10}$,$\sqrt{10})$ and $(-\sqrt{10}$,$-\sqrt{10})$.
Also, for the other question $(3,0)$ cannot be the answer as it does not satisfy the first equation.
