This isn't homework exactly, but rather I'm teaching a class on precalculus and I'm a little stuck on a question from the textbook.
Convert $x^2 + y^2 = 2ax$ to a polar equation, where $a>0$.
This is incredibly simple to do: note that $x^2 + y^2 = r^2$, and $x = r\cos\theta$. Then the equation is simply: $$r^2 = 2ar\cos\theta.$$ However, the answer in the back of the textbook is $$r = 2a\cos\theta.$$
Here's my question: Why can we divide by $r$? What happens if $r=0$? Not for every $\theta$ will $r=0$, so why is it valid to simply divide by it?
$r$ is the equation of a circle with radius $a$ whose center lies on the $x$-axis, and graphically the pole $r=0$ exists on the graph. So what gives?
Well, I admit they do not give enough detail. $$ (x-a)^2 + y^2 = a^2. $$ With $a > 0,$ this is tangent to the vertical axis at the origin. The polar expression $r = 2 a \cos \theta$ covers the circle twice, once with $ r$ positive while $\; - \pi/2 < \theta < \pi / 2,$ then again with $r$ negative while, say, $\; \pi/2 < \theta < 3 \pi / 2.$ At the odd multiples of $\pi/2,$ we get the origin because $r=0.$
I guess you need to decide whether your students will be able to deal with negative $r.$