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

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Answer 1 · 2018-05-03T00:35:27

The conversion from Rectangular to Polar:
#x=rcostheta#
#y=rsintheta#

Substitute for #x# and #y#:
#9=(rcostheta-3)^2+(2rsintheta-9)^2#

#r^2cos^2theta-6rcostheta+9+4r^2sin^2theta-36sin^2theta+81=9#

#r^2cos^2theta-6rcostheta+4r^2sin^2theta-36rsintheta=-81#

#r(rcos^2theta-6costheta+4rsin^2theta-36sintheta)=-81#

#r=-81#

Or the more meaningful solution:

#rcos^2theta-6costheta+4rsin^2theta-36sintheta=-81#

#rcos^2theta+4rsin^2theta=-81+6costheta+36sintheta#

#r(cos^2theta+4sin^2theta)=-81+6costheta+36sintheta#

#r=(-81+6costheta+36sintheta)/(cos^2theta+4sin^2theta)#