# Can you help me Transform from a polar to a rectangular equation?

## $r = 2 \sin \theta + 2 \cos \theta$

Mar 8, 2018

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

#### Explanation:

We use the following three definitions to convert between Cartesian and polar coordinates

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

$x = r \cos \theta \Rightarrow \cos \theta = \frac{x}{r}$

$y = r \sin \theta \Rightarrow \sin \theta = \frac{y}{r}$

Using these, we get the equation as

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

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

$\Rightarrow {x}^{2} - 2 x + {y}^{2} - 2 y = 0$

We now complete the square for both $x$ and $y$ to get the eqn in standard form

${\left(x - 1\right)}^{2} - 1 + {\left(y - 1\right)}^{2} - 1 = 0$

So finally,

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

Mar 8, 2018

Given:

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

Please observe the graph of the polar equation:

Multiply both sides by $r$:

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

Substitute ${r}^{2} = {x}^{2} + {y}^{2} , r \sin \left(\theta\right) = y , \mathmr{and} r \cos \left(\theta\right) = x$:

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

Subtract $2 y + 2 x$ from both sides:

${x}^{2} - 2 x + {y}^{2} - 2 y = 0$

Add ${h}^{2} + {k}^{2}$ to both dies:

${x}^{2} - 2 x + {h}^{2} + {y}^{2} - 2 y + {k}^{2} = {h}^{2} + {k}^{2}$

Using the pattern ${\left(x - h\right)}^{2} = {x}^{2} + 2 h x + {h}^{2}$, we observe that $h = 1$

${x}^{2} - 2 x + {1}^{2} + {y}^{2} - 2 y + {k}^{2} = {1}^{2} + {k}^{2}$

Collapse the left side into a square:

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

Using the pattern ${\left(y - k\right)}^{2} = {y}^{2} + 2 k x + {k}^{2}$, we observe that $k = 1$

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

Collapse the left side into a square and combine like terms on the right:

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

Write in standard form:

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

This is a circle with its center at the point $\left(1 , 1\right)$ and it radius, $r = \sqrt{2}$:

Please observe that the graph is identical to the original equation: