How do you convert $(x-r)^2 + y^2 = r^2$ to polar form? Why do I get $0 = 0$?

Question

I am trying to convert this circle from cartesian to polar form. I have googled and searched for various guides/youtube videos and followed them. The answer is supposed to be in polar form $r = 2 \cos \theta$

After drawing the triangle, I could see that:

$x-r = r \cos \theta$

$y = r \sin \theta$

I had tried expanding it this way:

$(x-r)^2 + y^2 = r^2$ // equation of the circle

$x^2 - 2xr + r^2 + y^2 = r^2$

$x^2 - 2xr + y^2 = 0$

$(r \cos \theta + r)^2 - r^2\sin^2 \theta = 0$

$r^2\cos^2\theta + 2r^2\cos\theta + r^2 - 2r^2\cos\theta - 2r^2 + r^2\sin^2\theta = 0 $

The 2nd the 4th term cancels each other out

$r^2\cos^2\theta + r^2 - 2r^2 + r^2\sin^2\theta = 0 $

$r^2(\cos^2\theta + \sin^2\theta) - r^2 = 0$

$r^2 - r^2 = 0$

$0 = 0$ ?

What am I doing wrong?

Answer 1

Using a different named constant to prevent confusion we have \begin{align} (x-a)^2+y^2=a^2 &\iff(r\cos{(\theta)}-a)^2+(r\sin{(\theta)})^2=a^2\\ &\iff r^2(\sin^2{(\theta)}+\cos^2{(\theta)})-2ar\cos{(\theta)}+a^2=a^2\\ &\iff r^2-2ar\cos{(\theta)}=0\\ &\iff r(r-2a\cos{(\theta)})=0\\ &\iff r=0\text{ or }r=2a\cos{(\theta)} \end{align}

Answer 2

$(x-r)^{2}+y^{2}=r^2$ is a circle centred at $(r,0)$ with radius $r$ (with $r$ constant). As mentioned in the comments, this $r$ is not the same $r$ used in the transformation to polar coordinates.

So using the polar coordinates transformations $x=r_{1}cos(\theta)$ and $x=r_{1}sin(\theta)$ gives $(x-r)^{2}+y^{2}=x^{2}-2xr+r^{2}+y^{2}=r_{1}^{2}(cos^{2}(\theta)+sin^{2}(\theta))-2rr_{1}cos(\theta)+r^{2}=r_{1}(r_{1}-2rcos(\theta))+r^{2}=r^{2}.$

Or $r_{1}(r_{1}-2rcos(\theta))=0$, so we have $r_{1}=0$ which gives the origin and $r_{1}=2rcos(\theta)$ which gives the required circle.