# Cartesian to polar

1. Dec 7, 2009

### Calcgeek123

1. The problem statement, all variables and given/known data
Establish an equation in polar coordinates for the curve x^2+y^2=4y-2x

2. Relevant equations

n/a

3. The attempt at a solution
I know that x^2+y^2=r^2 so I used substitution, and now have r^2=4y-2x. Now this next part, I'm really not sure if I'm allowed to do this... i know that x=rcos@ (let @=theta) and y=rsin@. So i simply substituted to get r^2=4rsin@-2cos@. Is that allowed? Also, i feel like my answer should just look like r=..... So should I take the sqaure root of both sides?
Thank you, very appreciated.

2. Dec 7, 2009

### Staff: Mentor

You should have r^2 = 4rsin@ - 2rcos@. You lost an "r" on your cosine term. You can simplify by dividing both sides by r, but you don't want to take the square root of both sides. You normally have to be careful of dividing by r, in case r happens to be equal to zero. That's not a problem in this case, since the curve doesn't go through the origin.

3. Dec 7, 2009

### Calcgeek123

Yeah, the missing r is a type. Dividing makes sense though, duh. thanks so much!