## Trigonometry (11th Edition) Clone

$(x-1)^2+y^2 = 1$ We can see the graph below:
$r = 2~cos~\theta$ We know that $r = \sqrt{x^2+y^2}$ Since $x = r~cos~\theta$, then $cos~\theta = \frac{x}{\sqrt{x^2+y^2}}$ We can find the rectangular coordinates: $r = 2~cos~\theta$ $\sqrt{x^2+y^2} = \frac{2x}{\sqrt{x^2+y^2}}$ $x^2+y^2 = 2x$ $x^2-2x+y^2 = 0$ $(x-1)^2+y^2 = 1$ We can see the graph below: