When do the curves $r=a(1+\sin\theta)$ $r=a(1-\sin\theta)$ intersect?

Question

By converting the equations to $x$- and $y$-components, and setting them equal, I get they intersect at $\theta=0,\pi$, giving the points $(a,0)$ and $(a,\pi)$. But I don't get the point $(0,0)$--how would I find that?

Answer 1

the third point of intersection happens when $\theta = 3\pi/2, r_1 = 0$ and $\theta = \pi/2, r_2 = 0.$ that is the point $r = 0$ in polar coordinates and $(0,0)$ in cartesian coordinates.

Answer 2

If you transform to Cartesian coordinates, you get

$$x^2+y^2=a\sqrt{x^2+y^2}+ay\\ x^2+y^2=a\sqrt{x^2+y^2}-ay$$

Equating these two equations gives you $x^2=a|x|$, which has three solutions $(0,0),(0,\pm a)$.