# How do you convert #r = 8 + 5cos(theta)# into cartesian form?

##### 1 Answer

#### Explanation:

Remember the conversions:

#x = rcostheta#

#y = rsintheta#

#x^2+y^2=r^2#

#+-sqrt(x^2+y^2)=r#

Note that

#r = 8 + 5costheta#

#r^2 = 8r +5rcostheta#

#x^2+y^2 = 8sqrt(x^2+y^2) + 5x#

Since we have

#x^2-5x+y^2 = 8sqrt(x^2+y^2)#

#(x^2-5x+y^2)^2 = 64(x^2+y^2)#

#x^4-10x^3+2x^2y^2+25x^2-10xy^2+y^4 = 64x^2+64y^2#

#x^4 - 10x^3 - 39x^2+2x^2y^2-10xy^2-64y^2+y^4 = 0#

We *could* try to simplify this in other ways, but this is a valid form of this equation, so we'll leave it like this.

*Final Answer*