Converting rectangular equation to polar equation How do I convert the following rectangular equation to a polar equation:
$$
(x- 3)^2 + (y+2)^2 = 1
$$
I was able to simplify it to the following:
$$
r^2 - 6r\cos(\theta) + 4r\sin(\theta) + 12 = 0
$$
but I am at a loss as to how to simplify further...
I'm trying to get r on one side and the answer on the other. In other words, r = ?
 A: You can combine $-6r\cos(\theta)+4r\sin(\theta)=2\sqrt{13}\sin(\theta-\arctan \frac 32)$   Whether that is a simplification I leave to you
Added:  from your edit, if you want $r=$ something, notice that you have a quadratic equation, so feed it to the quadratic formula.  $$r=(2 \sin \theta - 3 \cos \theta) \pm \sqrt{(2 \sin \theta - 3 \cos \theta)^2-12}$$
A: Hint: 
Make the following change of variables:
$$x = 3+r \cos\theta, \quad y = -2+r \sin \theta.$$
Cheers.
A: $r^2- 6r cos(\theta)+ 4r sin(\theta)+ 12= 0$
$r^2- (6 cos(\theta)+ 4 sin(\theta))r+ 12= 0$
is a quadratic equation in r.  It can be solved using the quadratic formula.
$r= \frac{4 sin(\theta)- 6cos(\theta)\pm\sqrt{(4 sin(\theta)- 6 cos(\theta))- 48}}{2}$
A: Given that you're trying to solve for $r$, threat the equation $r^2-6r\cos\theta+4r\sin\theta+12=0$ as a quadratic equation, where $r$ is the only variable and everything else is treated as constants.
But before we begin solving for $r$, it would be best to take the terms $-6r\cos\theta$ and $4r\sin\theta$ and factor out the $r$. This will help make the actual solving part a little easier.
$r^2-6r\cos\theta+4r\sin\theta+12=0$
$r^2+\left(4\sin\theta-6\cos\theta\right)r+12=0$
Note that in addition to factoring out $r$, I also reversed the order of the 2 terms so that the positive term is followed by the negative term, though this step isn't necessarily mandatory.
From here we can either use Completing the Square or Quadratic Formula to solve the equation for $r$. I'll be using Completing the Square.
$r^2+\left(4\sin\theta-6\cos\theta\right)r+12=0$
$r^2+\left(4\sin\theta-6\cos\theta\right)r=-12$
$r^2+\left(4\sin\theta-6\cos\theta\right)r+\left(2\sin\theta-3\cos\theta\right)^2=-12+\left(2\sin\theta-3\cos\theta\right)^2$
$\left(r+2\sin\theta-3\cos\theta\right)^2=\left(2\sin\theta-3\cos\theta\right)^2-12$
$r+2\sin\theta-3\cos\theta=\pm\sqrt{\left(2\sin\theta-3\cos\theta\right)^2-12}$
$r=-2\sin\theta+3\cos\theta\pm\sqrt{\left(2\sin\theta-3\cos\theta\right)^2-12}$
$r=3\cos\theta-2\sin\theta\pm\sqrt{\left(2\sin\theta-3\cos\theta\right)^2-12}$
