How do you convert $9=(4x+7)^2+(-y+7)^2$ into polar form?

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Answer 1 · 2016-04-15T18:52:46

$16r^2cos^2theta+r^2sin^2theta+56rcostheta-14rsintheta+89=0$

$9=16x^2+56x+49+y^2-14y+49$

$0=16x^2+56x+y^2-14y+49+49-9$

$0=16x^2+y^2+56x-14y+89$

Now use the formulas
$x=rcos theta and y=rsintheta$

$0=16r^2cos^2theta+r^2sin^2theta+56rcostheta-14rsintheta+89$

How do you convert 9=(4x+7)^2+(-y+7)^29=(4x+7)^2+(-y+7)^2 into polar form?