How do you convert r = 1-2 cosθ into rectangular forms?

Jan 24, 2016

${\left({x}^{2} + {y}^{2} - b x\right)}^{2} = {a}^{2} \left({x}^{2} + {y}^{2}\right)$
a = 1; b = -2
${\left({x}^{2} + {y}^{2} + 2 x\right)}^{2} = \left({x}^{2} + {y}^{2}\right)$

Explanation:

use the Polar to Cartesian transformation:
${r}^{2} = {x}^{2} + {y}^{2}$
$x = r \cos \theta$

$r = a + b \cos \theta$
${r}^{2} = a r + r b \cos \theta$
${r}^{2} - r b \cos \theta = a r$
${x}^{2} + {y}^{2} - b x = a r$ square the entire thing and write
${\left({x}^{2} + {y}^{2} - b x\right)}^{2} = {\left(a r\right)}^{2}$
${\left({x}^{2} + {y}^{2} - b x\right)}^{2} = {a}^{2} \left({x}^{2} + {y}^{2}\right)$

a = 1; b = -2
${\left({x}^{2} + {y}^{2} + 2 x\right)}^{2} = \left({x}^{2} + {y}^{2}\right)$