How do you convert $4=(x+8)^2+(y-6)^2$ into polar form?

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Answer 1 · 2016-05-23T09:04:06

Polar form is $r^2+4r(4costheta-3sintheta)+96=0$

A Cartesian point $(x,y)$ in polar form is $(r,theta)$ , where

$x=rcostheta$ and $y=rsintheta$ and hence

$x^2+y^2=r^2cos^2theta+r^2sin^2theta=r^2$

Hence $4=(x+8)^2+(y-6)^2$ can be written as

$(rcostheta+8)^2+(rsintheta-6)^2=4$

or $r^2cos^2theta+16rcostheta+64+r^2sin^2theta-12rsintheta+36=4$

or $r^2+r(16costheta-12sintheta)+64+36-4=0$

or $r^2+4r(4costheta-3sintheta)+96=0$

How do you convert 4=(x+8)^2+(y-6)^24=(x+8)^2+(y-6)^2 into polar form?