Convert polar equation to rectangular when there is a floating constant.

Question

I know how to convert a polar equation to a rectangular equation when it's in a form such as $$r=3\cos{\theta}$$ In that case, I would multiply both sides by $r$ to get $$r^2=3r\cos{\theta}$$ and then substitute to get $$x^2+y^2=3x$$ But what about when there's a floating constant, such as $$r=3-3\cos{\theta}$$ If I try to use the same method, I get $$r^2=3r-3r\cos{\theta}$$ and substitute to get $$x^2+y^2=\pm3\sqrt{x^2+y^2}-3x$$ Would that be correct?

Answer 1

Yes, you are correct, except for a possible simplification. You can drop the negative version since $$x^2+y^2=-3\sqrt{x^2+y^2}-3x$$ $$x^2+y^2=-3(\sqrt{x^2+y^2}+x)$$ The left side is always nonnegative. The right side will be negative unless two conditions are met.

1. $x\le0$

2. $|x|\ge\sqrt{x^2+y^2}$

And since $|x|=\sqrt{x^2}$ and $y^2\ge0$, the second condition can only be true if $y=0$. This yields the solution point $(0,0)$, but the positive version already gives that solution point. This makes the final answer: $$x^2+y^2=3\sqrt{x^2+y^2}-3x$$

The other option to get rid of the $\pm$ is to move the $3x$ to the left and then square both sides of the equation, giving: $$(x^2+y^2+3x)^2=9(x^2+y^2)$$

Answer 2

I believe you have worked enough without the floating constant. Trick here is to rope in $r$ by multiplication.

$$r=3-3\cos{\theta};~ r=1+2\sin{\theta}$$

$$r^2=3r-3r\cos{\theta};~ r^2=r+2 r\sin{\theta}$$

$$x^2+y^2=3\sqrt{x^2+y^2}-3 x ;~ x^2+y^2=\sqrt{x^2+y^2}+2 y ;$$ and so on..