Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 8 - Complex Numbers, Polar Equations, and Parametric Equations - Section 8.5 Polar Equations and Graphs - 8.5 Exercises - Page 395: 62


$(x-1)^2+y^2 = 1$ We can see the graph below:

Work Step by Step

$r = 2~cos~\theta$ 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:
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