Question

How do you convert `4=(x-4)^2+(y-1)^2` into polar form?

Answer 1

`r^2 - 8r \cos \theta - 2 r \sin \theta + 13 = 0`

Explanation:

There is a standard way to convert from rectangular coordinates to polar coordinates: replace all instances of `x` with `r\cos \theta` and all instances of `y` with `r \sin \theta`. The equation of the curve (a circle) can be expanded and written as follows:

`x^2 + y^2 - 8x - 2y + 13 = 0`

Now do all the replacement work.

`(r \cos \theta)^2 + (r \sin \theta)^2 - 8 r \cos \theta - 2 r \ sin \theta + 13 = 0`

`\Rightarrow r^2 - 8 r \cos \theta - 2 r \sin \theta + 13 = 0`