in polar this is
#vec r = 3 cos theta \ hat r#
but #hat r = ((cos theta), (sin theta))# in Cartesian
so in Cartesian #vec r = 3 cos theta ((cos theta), (sin theta)) = 3 ((cos^2 theta), (sin theta cos theta))#
#= 3 ((1/2( 1 + cos 2 theta)), (1/2 sin 2 theta )) = 3/2 (( 1 + cos 2 theta), ( sin 2 theta )) = ((x),(y))#
so if you want to param in, say, # t#, you could if you like sub #t = 2 theta# to get
#x = 3/2( 1 + cos t), y = 3/2 sin t#
you lose a bit of information doing that so I'd go with
#x = 3/2( 1 + cos 2t), y = 3/2 sin 2t#
from this you can extricate:
#2/3 x - 1 = cos 2t, 2/3 y = sin 2t# such that
#(2/3 x - 1)^2 +( 2/3 y)^2 = 1#
#( x - 3/2)^2 + y^2 = (3/2)^2# which is a Cartesian circle of radius #3/2# and centre #(3/2, 0)#
