How do you convert #4=(x-3)^2+(y-5)^2# into polar form?

1 Answer
George C.
May 2, 2016

#r^2 - 2r(3 cos theta + 5 sin theta) + 30 = 0#

Explanation:

To convert into polar form, substitute:

#{ (x = r cos theta), (y = r sin theta) :}#

#4 = (r cos theta - 3)^2+(r sin theta - 5)^2#

#=r^2 cos^2 theta - 6r cos theta + 9 + r^2 sin^2 theta - 10r sin theta + 25#

#=r^2 (cos^2 theta + sin^2 theta) - 2r(3cos theta + 5 sin theta)+34#

#=r^2 - 2r(3 cos theta + 5 sin theta) + 34#

Subtract #4# from both ends to get:

#r^2 - 2r(3 cos theta + 5 sin theta) + 30 = 0#

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