Question

How do you convert `(x-2)^2 + (y-3)^2 = 1` into a polar equation?

Answer 1

`r^2-2(2 cos(theta)+3 sin(theta))r+12=0`

Explanation:

The pass relationships `x=r cos(theta),y=r sin(theta)` substituted in the cartesian circunference representation, gives:
`(r cos(theta)-2)^2+(r sin(theta)-3)^2=1`. Simplifiying
`r^2-2(2 cos(theta)+3 sin(theta))r+12=0`