How do you convert r=1/2costheta into cartesian form?

Jun 21, 2016

${\left(x - \frac{1}{4}\right)}^{2} + {y}^{2} = {\left(\frac{1}{4}\right)}^{2}$

Explanation:

the basic ideas are $x = r \cos \theta , y = r \sin \theta$ so ${x}^{2} + {y}^{2} = {r}^{2} {\cos}^{2} \theta + {r}^{2} {\sin}^{2} \theta = {r}^{2}$

so using $r = \sqrt{{x}^{2} + {y}^{2}}$ and $\cos \theta = \frac{x}{r} = \frac{x}{\sqrt{{x}^{2} + {y}^{2}}}$ you can plug and play

So $\sqrt{{x}^{2} + {y}^{2}} = \frac{1}{2} \frac{x}{\sqrt{{x}^{2} + {y}^{2}}}$
${x}^{2} + {y}^{2} = \frac{x}{2}$
${x}^{2} - \frac{x}{2} + {y}^{2} = 0$
completing the square in x
${\left(x - \frac{1}{4}\right)}^{2} - \frac{1}{16} + {y}^{2} = 0$
${\left(x - \frac{1}{4}\right)}^{2} + {y}^{2} = {\left(\frac{1}{4}\right)}^{2}$

that's a circle centred on (1/4, 0) radius 1/4.