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:

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

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

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

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

How do you convert 9=(x-3)^2+(2y-9)^29=(x-3)^2+(2y-9)^2 into polar form?