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

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

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

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

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

How do you convert 1=(-x+4)^2+(2y+9)^21=(-x+4)^2+(2y+9)^2 into polar form?