Rewrite the polar equation r = 3 cos(theta) as a Cartesian equation.

Let me think about this

$$\\x^2+y^2=r^2\\ r=3cos\theta\\ x=rcos\theta\\ r^2=3rcos\theta\\ so\\ r^2=3x\\ so\\ x^2+y^2=3x\\ x^2-3x+1.5^2+y^2=0+2.5^2\\ (x-1.5)^2+y^2=(1.5)^2\\ $ This is a circle centre (1.5,\;0) and radius 1.5$$ $

NOW I HAVE GOT IT   !!

Thanks Chris  :))

r = 3cosΘ     Multiply both sides by r

r^2  = 3rcosΘ

x^2 + y^2  = 3x

Here is the graph of each form.......notice that they overlap perfectly......https://www.desmos.com/calculator/hg8gmfwwu2

Thanks CPhill,

I am trying to get my head around this solution.

I can see how you have done it but is do not intrisically understand ://

Melody....the reason for multiplying by "r" on each side is that we have no Cartesian equivalent for cosΘ...but rcosΘ is easily transformed to "x"....also, multiplying the left side by "r" gets rid of a "root" problem since r = √[x² + y²]...but.....r² = x² + y².....and the root "disappears".....