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

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Answer 1 · 2016-10-04T09:13:48

$r=6costheta$

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

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

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

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

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

or $r^2-6rcostheta=0$

or $r^2=6rcostheta$

or $r=6costheta$

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