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

1 Answer
Dec 18, 2015

# 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#