How do you change the rectangular equation $x^{2} + y^{2} + 8 x = 0$ $x^2+y^2+8x=0$ into polar form?

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Answer 1 · 2016-11-03T05:37:44

$r + 8 cos θ = 0$ $r+8costheta=0$

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

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

Hence $x^{2} + y^{2} + 8 x = 0$ $x^2+y^2+8x=0$

$\Leftrightarrow r^{2} + 8 r cos θ = 0$ $hArrr^2+8rcostheta=0$

or $r + 8 cos θ = 0$ $r+8costheta=0$

How do you change the rectangular equation x^2+y^2+8x=0x2+y2+8x=0x^2+y^2+8x=0 into polar form?