Intersection of rectangular and polar equations

Question

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 $(3,0) and (-3,0)$...

Answer 1

$x = r\cos\theta, \ y= r\sin\theta$.

$\theta = - 3\pi/4$.

$r = (\sqrt{20})$.

Combining:

$x = (\sqrt{20})\cos(-3\pi/4)$ ;

$y = (\sqrt{20})\sin(-3\pi/4)$ .

$\cos(-3\pi/4) = \cos(3\pi/4) =$

$- (1/2)\sqrt{2}$.

$\sin(-3\pi/4) = - \sin(3\pi/4) =$

$ -(1/2)\sqrt{2}$.

Finally:

$x = -(\sqrt{20})/(\sqrt{2}) = - (\sqrt{10})$;

$y = - (\sqrt{20})/(\sqrt{2})= - (\sqrt{10})$;

For the given $\theta$ value only one point of intersection (in the third quadrant).

Answer 2

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.