How do you convert $r=6cosθ$ into a cartesian equation?

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Answer 1 · 2015-08-16T21:17:26

I found the equation: $x^2+y^2-6x=0$ which represents a circle.

Consider the connection between Cartesian and Polar:
$r=sqrt(x^2+y^2)$
$theta=arcrtan(y/x)$

And:

$x=rcos(theta)$
$y=rsin(theta)$

In your case you get:
$sqrt(x^2+y^2)=6x/r$
$sqrt(x^2+y^2)=6x/sqrt(x^2+y^2)$
$x^2+y^2-6x=0$

How do you convert r=6cosθr=6cosθ into a cartesian equation?