4 Answers

Use:
r = √(x^2 + y^2) and cosθ = x/r.
So we have:
r = 4cosθ
==> √(x^2 + y^2) = 4x/r, by substituting
==> √(x^2 + y^2) = 4x/√(x^2 + y^2), since r = √(x^2 + y^2)
==> x^2 + y^2 = 4x
==> (x^2 + 4x) + y^2 = 0
==> (x^2 + 4x + 4) + y^2 = 4, by completing the square
==> (x + 2)^2 + y^2 = 4 = 2^2.
This is a circle with a center of (2, 0) and a radius of 2.
I hope this helps!

if r = 4cosÎ¸
r^2 = 4rcosÎ¸
now r^2 = x^2 + y^2
and x = rcosÎ¸, so we have
x^2 + y^2 = 4x
thus
x^2 + 4x + 4 + y^2 = 4
or
(x + 2)^2 + y^2 = 4
which is a circle centered at (2, 0) of radius 2.

r = 4cosÎ¸
rÂ² = 4r cosÎ¸
Does that help? See if you can show that this is a circle, centre (2, 0), radius 2.

rectangular coordinates (rcos Î¸, rsinÎ¸) = (4cos^2Î¸,4sinÎ¸cosÎ¸) = 2 {(cos2Î¸+1), sin(2Î¸)}