How do you convert #4=(x+8)^2+(y-6)^2# into polar form?

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Answer 1 · 2016-05-23T09:04:06

Polar form is #r^2+4r(4costheta-3sintheta)+96=0#

A Cartesian point #(x,y)# in polar form is #(r,theta)#, where

#x=rcostheta# and #y=rsintheta# and hence

#x^2+y^2=r^2cos^2theta+r^2sin^2theta=r^2#

Hence #4=(x+8)^2+(y-6)^2# can be written as

#(rcostheta+8)^2+(rsintheta-6)^2=4#

or #r^2cos^2theta+16rcostheta+64+r^2sin^2theta-12rsintheta+36=4#

or #r^2+r(16costheta-12sintheta)+64+36-4=0#

or #r^2+4r(4costheta-3sintheta)+96=0#