How do you convert $9=(-2x+y)^2-5y+3x$ into polar form?

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Answer 1 · 2018-03-14T14:57:17

$9=4r^2cos^2(theta)-4r^2sinthetacostheta+r^2sin^2(theta)-5rsintheta+3rcostheta=r(sintheta(r(sintheta−4costheta)−5)+costheta(4rcostheta+3))$

$x=rcostheta$
$y=rsintheta$

$9=(-2(rcostheta)+rsintheta)^2-5rsintheta+3rcostheta$

$9=4r^2cos^2(theta)-4r^2sinthetacostheta+r^2sin^2(theta)-5rsintheta+3rcostheta$

$9=r(sintheta(r(sintheta−4costheta)−5)+costheta(4rcostheta+3))$

How do you convert 9=(-2x+y)^2-5y+3x9=(-2x+y)^2-5y+3x into polar form?