How do you convert #9=(x+3)^2+(y+8)^2# into polar form?

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Answer 1 · 2016-05-22T12:54:54

#9=(x+3)^2+(y+8)^2# in polar form can be written as

#r^2+2r(3costheta+8sintheta)+64=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 #9=(x+3)^2+(y+8)^2# can be written as

#(rcostheta+3)^2+(rsintheta+8)^2=9#

or #r^2cos^2theta+6rcostheta+9+r^2sin^2theta+16rsintheta+64=9#

or #r^2+r(6costheta+16sintheta)+64=0#

or #r^2+2r(3costheta+8sintheta)+64=0#