How do you convert #9=(4x+7)^2+(6y-9)^2# into polar form?

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Answer 1 · 2016-05-19T17:31:41

#r^2(4+5sin^2theta)+r(14costheta-27sintheta)+18=0#

The relation between polar coordinates #(r,theta)# and Cartesian coordinates #(x,y)# is given by

#x=rcostheta#, #y=rsintheta#, #r^2=x^2+y^2#.

Using them #(4x+7)^2+(6y-9)^2=9# is

#(4x+7)^2+(6y-9)^2=9#

#(4rcostheta+7)^2+(6rsintheta-9)^2=9#

or #16r^2cos^2theta+56rcostheta+36r^2sin^2theta-108rsintheta+81=9#

or #16r^2+56rcostheta+20r^2sin^2theta-108rsintheta+72=0#

or #4r^2+14rcostheta+5r^2sin^2theta-27rsintheta+18=0#

or #r^2(4+5sin^2theta)+r(14costheta-27sintheta)+18=0#