Graphing $(x^{2}+y^{2})^{3}=4x^{2}y^{2}$: when is cancelling a function, $r(\theta)$ in parametric equation valid

Question

I am solving a question that asks me to graph $$(x^{2}+y^{2})^{3}=4x^{2}y^{2}.$$ I convert this into polar coordinates equation, to get $$(r^{2})^{3}=4(r \cos\theta)^{2}(r\sin\theta)^{2}\Rightarrow r^{6}=r^{4}\sin^2(2\theta).$$ I wonder if I can cancel $r^{4}$ to further simplify it; the book actually did it. But I am not sure when/why doing this is valid? Is it because $``r$ is a function that is not identically zero over its domain$"$?

Hope someone can help answer this.

Answer 1

You can't immediately cancel $r^4$. Note that $$r^6=r^4\sin^22\theta\implies r^4(r^2-\sin^22\theta)=0.$$ The case $r^4=0$ yields $r=0$ which in turn gives $(x,y)=(0,0)$. Notice that this is indeed a (trivial) solution to the original equation.

You can then proceed to consider the other case $r=\pm\sin2\theta$.