Intersection of two circles. Let $C_1$ and $C_2$ be the circles: $\rho=a\sin\theta, \rho=a(\cos\theta + \sin\theta)$ respectively. The graphs of these two circles are 
From the graphs, we see that the intersection points are $(0,0)$, $(\pi/2, a)$. But when we solve the system of equations: $\rho=a\sin\theta, \rho=a(\cos\theta + \sin\theta)$, we obtain $(\theta, \rho)=(\pi/2, a)$ or $(-\pi/2, -a)$. $(\pi/2, a)$, $(-\pi/2, -a)$ are different from $(0,0)$, $(\pi/2, a)$. I am confused. Thank you very much.
 A: Note that $\left(-\frac{\pi}{2},-a\right)$ is another representation of the point $\left(\frac{\pi}{2},a\right)$ (to see this, draw the radius $-a$ at an angle of $-\frac{\pi}{2}$). So the two points you get are actually the same point written differently.
As for $(0,0)$, note that $a\sin\theta$ passes through $(0,0)$ when $\theta$ is a multiple of $\pi$, while $a(\cos\theta + \sin\theta)$ passes through $(0,0)$ when $\theta$ is $\frac{3\pi}{4}\pm n\pi$ where $n$ is an integer. Thus the two curves never pass through $(0,0)$ at the same value of $\theta$, which is why you don't see this solution.
A: The issue here is that one Cartesian point has multiple polar representations.
The two points that you found, $(\pi/2,a)$ and $(-\pi/2,-a)$, both refer to to same location in Cartesian space. Also, you did not find an intersection at the origin because the two circles reach it with different values of $\theta$.
Assuming $a\ne0$, $a\sin\theta=0$ implies that $\theta=0$ or $\theta=\pi$, for $0\le\theta<2\pi$. For the second circle, $a(\cos\theta+\sin\theta)=0$ gives $\theta=3\pi/4$ or $\theta=7\pi/4$ for $0\le\theta<2\pi$. That means that $C_1$ contains the points $(0,0)$ and $(\pi,0)$, while $C_2$ contains the points $(3\pi/4,0)$ and $(7\pi/4,0)$. While these all represent the Cartesian point $(0,0)$, they are not the same in polar form. Whether or not you can consider this to be an intersection depends on the context of the problem.
