Answer
(a) See the graph.
(b) $4(x^2+y^2)=(x-4)^2$
Work Step by Step
(b)
$x=r~cos~θ$
$y=r~sin~θ$
$r^2=x^2+y^2$
$r=\frac{4}{2+cos~θ}$
$r=\frac{4}{2+\frac{x}{r}}$
$2r+x=4$
$2r=4-x~~$ (Square both sides)
$4r^2=(4-x)^2$
$4(x^2+y^2)=(x-4)^2$
