How do you convert the rectangular equation $y^{2} + {(x - 3)}^{2} = 9$ $y^2+(x-3)^2=9$ into polar form?

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Answer 1 · 2016-10-04T09:13:48

$r = 6 cos θ$ $r=6costheta$

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 $x^{2} + y^{2} = r^{2}$ $x^2+y^2=r^2$

Hence $y^{2} + {(x - 3)}^{2} = 9$ $y^2+(x-3)^2=9$ can be rewritten as

$y^{2} + x^{2} + 9 - 6 x = 9$ $y^2+x^2+9-6x=9$

or $y^{2} + x^{2} - 6 x = 0$ $y^2+x^2-6x=0$

or $r^{2} - 6 r cos θ = 0$ $r^2-6rcostheta=0$

or $r^{2} = 6 r cos θ$ $r^2=6rcostheta$

or $r = 6 cos θ$ $r=6costheta$

How do you convert the rectangular equation y^2+(x-3)^2=9y2+(x−3)2=9y^2+(x-3)^2=9 into polar form?