# What is the cartesian equation for the polar curve  r = 2sin theta - 4cos theta ?

Jul 7, 2017

${x}^{2} + {y}^{2} = 2 y \pm 4 \sqrt{{x}^{2} + {y}^{2}}$

#### Explanation:

Use the conversions:

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

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

$r \sin \theta = y$

$r \cos \theta = x$

First, let's multiply both sides of the equation by $r$.

$r \cdot r = r \cdot \left(2 \sin \theta - 4\right)$

${r}^{2} = 2 r \sin \theta - 4 r$

Now we can substitute the rectangular forms.

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

${x}^{2} + {y}^{2} = 2 y \pm 4 \sqrt{{x}^{2} + {y}^{2}}$

We can simplify this further, but this is a good stopping point.

Jul 10, 2017

The cartesian rectangular equation is:

${x}^{2} + 4 x + {y}^{2} - 2 y = 0$

Which can also be written as:

${\left(x + 2\right)}^{2} + {\left(y - 1\right)}^{2} = {\sqrt{5}}^{2}$

Which is a circle centred on $\left(- 2 , 1\right)$ with radius $\sqrt{5}$

#### Explanation:

Assuming the correct equation to be:

$r = 2 \sin \theta - 4 \cos \theta$

To convert to polar coordinates to cartesian rectangular form, using the relationships:

 {: (x = rcos theta, cos theta=x/r), (y = rsin theta, sin theta=x/r) :} } => r^2 = x^2+y^2

So we can write the polar equation as follows:

$r = 2 \sin \theta - 4 \cos \theta$
$\setminus = 2 \left(\frac{y}{r}\right) - 4 \left(\frac{x}{r}\right)$

Multiply by $r$ and we get:

${r}^{2} = 2 y - 4 x$

$\therefore {x}^{2} + {y}^{2} = 2 y - 4 x$

$\therefore {x}^{2} + 4 x + {y}^{2} - 2 y = 0$

Although this would be sufficient, we can analyze a little further by completing the square on both $x$ and $y$:

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

$\therefore {\left(x + 2\right)}^{2} - 4 + {\left(y - 1\right)}^{2} - 1 = 0$
$\therefore {\left(x + 2\right)}^{2} + {\left(y - 1\right)}^{2} = 5$

$\therefore {\left(x + 2\right)}^{2} + {\left(y - 1\right)}^{2} = {\sqrt{5}}^{2}$

Which is a circle centred on $\left(- 2 , 1\right)$ with radius $\sqrt{5}$

graph{x^2 + 4x + y^2 - 2y = 0 [-10, 10, -5, 5]}