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

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Answer 1 · 2017-01-06T07:12:43

$r^2+4r(costheta+sintheta)+64=0$

The relation between polar coordinates $(r,theta)$ and corresponding Cartesian coordinates $(x,y)$ is given by

$x=rcostheta$ , $y=rsintheta$ and $r^2=x^2+y^2$ .

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

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

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

or $r^2(cos^2theta+sin^2theta)+16rcostheta+4rsintheta+64=0$

or $r^2+4r(costheta+sintheta)+64=0$

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