How do you convert $x^2+4x+y^2+4y=0$ to polar form?

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Answer 1 · 2016-10-19T10:25:06

$r+4sqrt2cos(theta-pi/4)=0$

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

$x=rcostheta$ and $y=rsintheta$ or $r^2=x^2+y^2$

Hence $x^2+4x+y^2+4y=0$ can be written as

$x^2+y^2+4x+4y=0$

or $r^2+4rcostheta+4rsintheta=0$

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

or $r+4(costheta+sintheta)=0$

or $r+4sqrt2(costhetaxx1/sqrt2+sinthetaxx1/sqrt2)=0$

or $r+4sqrt2(costhetacos(pi/4)+sinthetasin(pi/4))=0$

or $r+4sqrt2cos(theta-pi/4)=0$

How do you convert x^2+4x+y^2+4y=0x^2+4x+y^2+4y=0 to polar form?