# How do you graph r = 12cos(theta)?

Nov 3, 2016

Set your compass to a radius of 6, put the center point at $\left(6 , 0\right)$, and draw a circle.

#### Explanation:

Multiply both sides of the equation by r:

${r}^{2} = 12 r \cos \left(\theta\right)$

Substitute ${x}^{2} + {y}^{2}$ for ${r}^{2}$ and $x$ for $r \cos \left(\theta\right)$

${x}^{2} + {y}^{2} = 12 x$

The standard form of this type of equation (a circle) is:

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

To put the equation in this form, we need to complete the square for both the x and y terms. The y term is easy, we merely subtract $0$ in the square:

${x}^{2} + {\left(y - 0\right)}^{2} = 12 x$

To complete the square for the x terms, we add $- 12 x + {h}^{2}$ both sides of the equation:

${x}^{2} - 12 x + {h}^{2} + {\left(y - 0\right)}^{2} = {h}^{2}$

We can use the pattern, ${\left(x - h\right)}^{2} = {x}^{2} - 2 h x + {h}^{2}$ to find the value of h:

${x}^{2} - 2 h x + {h}^{2} = {x}^{2} - 12 x + {h}^{2}$

$- 2 h x = - 12 x$

$h = 6$

Substitute 6 for h into the equation of the circle:

${x}^{2} - 12 x + {6}^{2} + {\left(y - 0\right)}^{2} = {6}^{2}$

We know that the first three terms are a perfect square with h = 6:

${\left(x - 6\right)}^{2} + {\left(y - 0\right)}^{2} = {6}^{2}$

This is a circle with a radius of 6 and a center point $\left(6 , 0\right)$