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

Aug 16, 2015

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

#### Explanation:

Consider the connection between Cartesian and Polar:
$r = \sqrt{{x}^{2} + {y}^{2}}$
$\theta = a r c r \tan \left(\frac{y}{x}\right)$

And:

$x = r \cos \left(\theta\right)$
$y = r \sin \left(\theta\right)$

$\sqrt{{x}^{2} + {y}^{2}} = 6 \frac{x}{r}$
$\sqrt{{x}^{2} + {y}^{2}} = 6 \frac{x}{\sqrt{{x}^{2} + {y}^{2}}}$
${x}^{2} + {y}^{2} - 6 x = 0$