Polar coordinates from rectangle

[Steel_City82]

Heres where I am struggling, I can't seem to change equations from rectangular to polar and vice versa

an example

x^2+y^2-2ax=0

heres what I got when I tried
r=2a cos theta
and that's a graph of a rose curve, I think, I am about 10% sure on that answer

heres an example of one I have no clue on

(x^2+y^2)(arctan(y/x))^2=a^2

heres what I am thinkin on this one

the x^2+y^2 can = r^2 and the arctan (y/x) can = theta
so you would have (r^2)(theta^2)=a^2

I don't know, I just can't get this

 

[Tide]

What, exactly, are you having doubts about? It looks fine to me.

 

[HallsofIvy]

x^2+y^2-2ax=0
Okay, obviously $x^2+ y^2= r^2$ and $2ax= 2ar cos(\theta)$ so the is $r^2- 2ar cos(\theta)= 0$ which you can write as $r^2= 2ar cos(\theta)$ and, as long as r is not 0, divide by r to get $r= 2a cos(\theta)$ as you have.

(x^2+y^2)(arctan(y/x))^2=a^2
Again $r^2= x^2+ y^2$ and, essentially by definition, $arctan(y/x)= \theta$ so this is simply [itex]r²\theta^2= a^2[/itex] as you have.
Assuming everything is positive, you can reduce that to $r\theta= a$.