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

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Answer 1 · 2016-10-19T10:25:06

$r + 4 \sqrt{2} cos (θ - \frac{π}{4}) = 0$ $r+4sqrt2cos(theta-pi/4)=0$

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

$x = r cos θ$ $x=rcostheta$ and $y = r sin θ$ $y=rsintheta$ or $r^{2} = x^{2} + y^{2}$ $r^2=x^2+y^2$

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

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

or $r^{2} + 4 r cos θ + 4 r sin θ = 0$ $r^2+4rcostheta+4rsintheta=0$

or $r^{2} + 4 r (cos θ + sin θ) = 0$ $r^2+4r(costheta+sintheta)=0$

or $r + 4 (cos θ + sin θ) = 0$ $r+4(costheta+sintheta)=0$

or $r + 4 \sqrt{2} (cos θ \times \frac{1}{\sqrt{2}} + sin θ \times \frac{1}{\sqrt{2}}) = 0$ $r+4sqrt2(costhetaxx1/sqrt2+sinthetaxx1/sqrt2)=0$

or $r + 4 \sqrt{2} (cos θ cos (\frac{π}{4}) + sin θ sin (\frac{π}{4})) = 0$ $r+4sqrt2(costhetacos(pi/4)+sinthetasin(pi/4))=0$

or $r + 4 \sqrt{2} cos (θ - \frac{π}{4}) = 0$ $r+4sqrt2cos(theta-pi/4)=0$

How do you convert x2+4x+y2+4y=0x^2+4x+y^2+4y=0 to polar form?