How do you convert #1=(-x+4)^2+(2y+9)^2# into polar form?

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Answer 1 · 2018-03-01T15:40:16

#r^2(cos^2theta+4sin^2theta)-r(4costheta-18sintheta)+40=0#

Given:
#1=(-x+4)^2+(2y+9)^2#

#1=(-x+4)^2+(2(y+9/2))^2#
#1=(-x+4)^2+4(y+9/2)^2#

#x=rcostheta#
#y=rsintheta#

Substituting
#1=(-rcostheta+4)^2+(2rsintheta+9)^2#

#1=r^2cos^2theta-8rcostheta+16+4r^2sin^2theta+36rsintheta+81#

#r^2(cos^2theta+4sin^2theta)+r(-8costheta+36sintheta)+81-1=0#

#2r^2(cos^2theta+4sin^2theta)-2r(4costheta-18sintheta)+2(40)=0#

#r^2(cos^2theta+4sin^2theta)-r(4costheta-18sintheta)+40=0#