# How do you convert r=2+cosθ into rectangular form?

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#### Explanation

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#### Explanation:

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1
Feb 9, 2018

$\textcolor{b l u e}{{x}^{4} + 2 {x}^{2} {y}^{2} - 2 {x}^{3} + {y}^{4} - 2 {y}^{2} x + {x}^{2} - 4 {x}^{2} - 4 {y}^{2} = 0}$

#### Explanation:

From the diagram we can see that point P has polar coordinates
$\left(r , \theta\right)$ and Cartesian coordinates $\left(x , y\right)$.

And $\textcolor{w h i t e}{88} x = r \cos \left(\theta\right)$ , $y = r \sin \left(\theta\right)$

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

Also:

By Pythagoras' Theorem :

${r}^{2} = {\left(r \cos \theta\right)}^{2} + {\left(r \sin \theta\right)}^{2}$

Since:

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

Then:

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

Using these ideas:

$r = 2 + \cos \left(\theta\right)$

Substitute: $\textcolor{w h i t e}{88} r = \sqrt{{x}^{2} + {y}^{2}}$

$\sqrt{{x}^{2} + {y}^{2}} = 2 + \cos \left(\theta\right)$

Substitute: $\textcolor{w h i t e}{88} \cos \left(\theta\right) = \frac{x}{r}$

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

Substitute fro $\boldsymbol{r}$ again:

$\sqrt{{x}^{2} + {y}^{2}} = 2 + \frac{x}{\sqrt{{x}^{2} + {y}^{2}}}$

Multiply by: $\sqrt{{x}^{2} + {y}^{2}}$

$\sqrt{{x}^{2} + {y}^{2}} \cdot \sqrt{{x}^{2} + {y}^{2}} = 2 \sqrt{{x}^{2} + {y}^{2}} + \frac{x \sqrt{{x}^{2} + {y}^{2}}}{\sqrt{{x}^{2} + {y}^{2}}}$

${x}^{2} + {y}^{2} = 2 \sqrt{{x}^{2} + {y}^{2}} + \frac{x \cancel{\sqrt{{x}^{2} + {y}^{2}}}}{\cancel{\sqrt{{x}^{2} + {y}^{2}}}}$

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

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

Squaring:

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

$\textcolor{b l u e}{{x}^{4} + 2 {x}^{2} {y}^{2} - 2 {x}^{3} + {y}^{4} - 2 {y}^{2} x + {x}^{2} - 4 {x}^{2} - 4 {y}^{2} = 0}$

I did not include the steps in expanding ${\left({x}^{2} + {y}^{2} - x\right)}^{2}$. This would make the answer too long.

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