# Convert the polar equation to rectangular form? r = 3/9 cos θ − 8 sin θ

Nov 30, 2016

#### Explanation:

Multiply both sides by r:

${r}^{2} = \frac{3}{9} r \cos \left(\theta\right) - 8 r \sin \left(\theta\right)$

Substitute ${x}^{2} + {y}^{2}$ for ${r}^{2}$:

${x}^{2} + {y}^{2} = \frac{3}{9} r \cos \left(\theta\right) - 8 r \sin \left(\theta\right)$

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

${x}^{2} + {y}^{2} = \frac{1}{3} x - 8 y$

Move the terms on the right to the left by adding $- \frac{1}{3} x + 8 y$ to both sides of the equation:

${x}^{2} - \frac{1}{3} x + {y}^{2} + 8 y = 0$

We are going to complete the squares so add ${h}^{2} + {k}^{2}$ to both sides of the equation:

${x}^{2} - \frac{1}{3} x + {h}^{2} + {y}^{2} + 8 y + {k}^{2} = {h}^{2} + {k}^{2}$

Set the middle term in the right side of the pattern ${\left(x - h\right)}^{2} = {x}^{2} - 2 h x + {h}^{2}$ equal to the corresponding term in the equation:

$- 2 h x = - \frac{1}{3} x$

Solve for h:

$h = \frac{1}{6}$

Substitute the left side of the pattern into the equation:

${\left(x - h\right)}^{2} + {y}^{2} + 8 y + {k}^{2} = {h}^{2} + {k}^{2}$

Substitute $\frac{1}{6}$ for every h:

${\left(x - \frac{1}{6}\right)}^{2} + {y}^{2} + 8 y + {k}^{2} = {\left(\frac{1}{6}\right)}^{2} + {k}^{2}$

Set the middle term in the right side of the pattern ${\left(y - k\right)}^{2} = {y}^{2} - 2 k y + {k}^{2}$ equal to the corresponding term in the equation:

$- 2 k y = 8 y$

Solve for k:

$k = - 4$

Substitute the left side of the pattern into the equation:

${\left(x - \frac{1}{6}\right)}^{2} + {\left(y - k\right)}^{2} = {\left(\frac{1}{6}\right)}^{2} + {k}^{2}$

Substitute -4 for every k:

${\left(x - \frac{1}{6}\right)}^{2} + {\left(y - - 4\right)}^{2} = {\left(\frac{1}{6}\right)}^{2} + {\left(- 4\right)}^{2}$

Combine the right side terms:

${\left(x - \frac{1}{6}\right)}^{2} + {\left(y - - 4\right)}^{2} = \frac{577}{36}$

Write the right side as a square:

${\left(x - \frac{1}{6}\right)}^{2} + {\left(y - - 4\right)}^{2} = {\left(\frac{\sqrt{577}}{6}\right)}^{2}$

Done!

This a circle with a radius of $\frac{\sqrt{577}}{6}$ centered at $\left(\frac{1}{6} , - 4\right)$