# How do you convert r = 8 + 5cos(theta) into cartesian form?

Jul 20, 2017

${x}^{4} - 10 {x}^{3} - 39 {x}^{2} + 2 {x}^{2} {y}^{2} - 10 x {y}^{2} - 64 {y}^{2} + {y}^{4} = 0$

#### Explanation:

Remember the conversions:

$x = r \cos \theta$

$y = r \sin \theta$

${x}^{2} + {y}^{2} = {r}^{2}$

$\pm \sqrt{{x}^{2} + {y}^{2}} = r$

Note that $r = 8 + 5 \cos \theta$ is always positive, so we don't need the $\pm$ sign in front of our square root when making that substitution

$r = 8 + 5 \cos \theta$

${r}^{2} = 8 r + 5 r \cos \theta$

${x}^{2} + {y}^{2} = 8 \sqrt{{x}^{2} + {y}^{2}} + 5 x$

Since we have $x$ and $y$ mixed together, we shouldn't try to separate them out and solve for $y$ by itself. Instead, let's try to get this equation into the form $f \left(x , y\right) = 0$.

${x}^{2} - 5 x + {y}^{2} = 8 \sqrt{{x}^{2} + {y}^{2}}$

${\left({x}^{2} - 5 x + {y}^{2}\right)}^{2} = 64 \left({x}^{2} + {y}^{2}\right)$

${x}^{4} - 10 {x}^{3} + 2 {x}^{2} {y}^{2} + 25 {x}^{2} - 10 x {y}^{2} + {y}^{4} = 64 {x}^{2} + 64 {y}^{2}$

${x}^{4} - 10 {x}^{3} - 39 {x}^{2} + 2 {x}^{2} {y}^{2} - 10 x {y}^{2} - 64 {y}^{2} + {y}^{4} = 0$

We could try to simplify this in other ways, but this is a valid form of this equation, so we'll leave it like this.