Can we divide both side by $r$ in polar equation?

Question

I was studying a Coordinate Geometry's chapter especially dedicated to $\mathit circles$. There I found the equation of a circle with center $\mathcal {(r_1, \theta_1)}$ and radius $\pmb a$ in polar coordinate system is $a^2= r^2+ r_1^{2} -2rr_{1}\cos(\theta-\theta_1)$. So far I have understood how we got here. Then there was a corollary stating-

Cor. 1. If the circle passes through the pole (which is the origin in rectangular coordinate system) then $r_1=a$ and the equation of the circle becomes- $$a^2= r^2+ a^{2} -2ra\cos(\theta-\theta_1) \\ \implies r^2 =2ra\cos(\theta-\theta_1) \\\implies \bbox[yellow] { r=2a\cos(\theta-\theta_1)}$$ In the highlighted section above they divided both sides of the equation by $r$. As far as I know, you cannot divide both sides by $r$ if $r$ happens to be $ 0$. But graphing the equation $r=2a\cos\left(\theta\right)$ where $\theta_1$ is set to be $0$ in desmos shows that it infact passes through the pole where $r=0$ when $\theta=90^{\circ}$.

$\mathcal {My ~Question:}$

So, is it legitimate to divide both sides by $r$ while $r$ can be equal to $0$ in graph of the equation? If so, why? I see in some textbooks they keep it in the former form without dividing by $r$ but in most place they divide both sides by $r$.

Answer 1

In fact, you can say :

"either $r=0$ (first curve) or $r=2a \cos(...)$.

The "or" between equations corresponds to "union" for the associated curves ; therefore you make the union of a point (the origin) and the second curve (which happens to contain already the origin...).

Finally, nothing is changed...

Answer 2

Strictly speaking $r^2 =rk$ implies $r=k$ or $r=0$

so $r^2 =2ra\cos(\theta-\theta_1)$ implies $r =2a\cos(\theta-\theta_1)$ or $r=0$

but since $\theta = \theta_1\pm \frac \pi2$ (or $\theta = \theta_1\pm 90^\circ$ if you are using degrees) also implies $r=0$, they may have felt adding $r=0$ explicitly was unnecessary