Question

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

Answer 1

Multiply both sides by `r` and substitute `r^{2}=x^{2}+y^{2}` and `rcos(t)=x` to get `x^{2}+y^{2}=5x`. Using the method of Completing the Square , this can be transformed to `(x-5/2)^{2}+y^{2}=25/4=(5/2)^{2}` to see that the curve is a circle of radius `5/2` centered at the point `(x,y)=(5/2,0)`.

Explanation:

Cartesian form means using rectangular coordinates rather than polar coordinates. Most typically, the "conversion equations" would be written with a `theta` rather than at `t` as:

`r^{2}=x^{2}+y^{2},\ x=rcos(theta),\ y=rsin(theta)`.

After the substitution mentioned above, the equation `x^{2}+y^[2}=5x` can be rewritten as `x^{2}-5x+y^{2}=0`, or `x^2-5x+(-5/2)^2+y^{2}=(-5/2)^2`, or

`(x-5/2)^{2}+y^{2}=(5/2)^{2}`