# How do you convert r = 5 cos (t) into cartesian form?

Nov 28, 2015

The equation relates $r \mathmr{and} \theta$, and describes a set of points $\left(r , \theta\right)$which form a curve in the plane. We want to know the curve's equation in $\left(x , y\right)$ coordinates.

#### Explanation:

You used $t$ instead of $\theta$. We can do that.

Start with knowing where we're going: the variables are related by things like
$x = r \cos \left(t\right) \mathmr{and} y = r \sin \left(t\right) \mathmr{and} {r}^{2} = {x}^{2} + {y}^{2.}$

You have $r = 5 \cos \left(t\right) ,$ let's create an r cos(t) by multiplying both sides by $r .$

${r}^{2} = 5 r \cos \left(t\right)$ . . . Now substitute and see that it turns into

${x}^{2} + {y}^{2} = 5 x .$

This is the answer. If you complete the square you get the form

${x}^{2} - 5 x + {y}^{2} = 0$ . . . Half of 5 is 5/2, whose square is 25/4:

${x}^{2} - 5 x + \frac{25}{4} + {y}^{2} = \frac{25}{4}$, or in circle form:

${\left(x - \frac{5}{2}\right)}^{2} + {\left(y - 0\right)}^{2} = {\left(\frac{5}{2}\right)}^{2}$,

You answer these ?'s: "It's a circle of center (?,?) and radius?."

...// dansmath strikes again! \\...