Question

# How do you find the polar equation for 3x^2+3y^2-2x=0?

How do you find the polar equation for 3x^2+3y^2-2x=0?

Answer 1

The equation is $3 r - 2 \cos \theta = 0$

#### Explanation:

To convert from cartesian coordinates $\left(x , y\right)$ to polar coordinates $\left(r , \theta\right)$ , we use the following

$x = r \cos \theta$

$y = r \sin \theta$

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

Our equation is

$3 {x}^{2} + 3 {y}^{2} - 2 x = 0$

$3 \left({x}^{2} + {y}^{2}\right) - 2 x = 0$

So,

$3 {r}^{2} - 2 r \cos \theta = 0$

$r \left(3 r - 2 \cos \theta\right) = 0$

$3 r - 2 \cos \theta = 0$

#### Explanation:

The relation between polar coordinates $\left(r , \theta\right)$ and Cartesian coordinates $\left(x , y\right)$ is given by

$x = r \cos \theta$ and $y = r \sin \theta$ i.e. $\tan \theta = \frac{y}{x}$ and ${r}^{2} = {x}^{2} + {y}^{2}$

Hence $3 {x}^{2} + 3 {y}^{2} - 2 x = 0 \Leftrightarrow 3 \left({x}^{2} + {y}^{2}\right) - 2 x = 0$

or $3 {r}^{2} - 2 r \cos \theta = 0$

or $3 r - 2 \cos \theta = 0$

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