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I'm trying to convert

$x^2 +y^2 =(2-x)^2$

into a polar equation in the form $r=f(\theta)$. The answer is apparently

$r=\frac{2}{1+cos(\theta)}$,

but I can't seem to get this.

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4 Answers 4

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The easiest way to remember the formulas for converting polar to rectangular coordinates and vice versa is to draw the right triangle at the origin with sides $x$ and $y$, hypotenuse $r$, and angle $\theta$. From there, it's easy to see that: $$x^2 + y^2 = r^2$$ $$x = r\cos\left(\theta\right)$$$$y = r\sin\left(\theta\right)$$

Using these equations to solve for $r$, $$x^2 + y^2 = (2-x)^2$$ $$r^2 = (2-x)^2$$ $$ r = 2-x$$ $$ r = 2 - r\cos\left(\theta\right)$$ $$ r + r\cos\left(\theta\right) = 2$$ $$ r(1 + \cos\left(\theta\right)) = 2$$ $$ r = \frac{2}{1 + \cos\left(\theta\right)}$$

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$$x^2 +y^2 =(2-x)^2\implies y^2+4x-4=0$$ Now, using $x=r \cos(t)$ and $y=r \sin(t)$, this lead to $$r^2 \sin^2(t)+4r \cos(t)-4=0$$ which is a quadratic equation in $r$.

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Just plug in $x = r \cos \theta, y = r \sin \theta$ and simplify. Namely,

$$ (r \cos \theta)^2 + (r \sin \theta)^2 = (2 - r \cos \theta)^2 \iff \\ r^2 = 4 - 4r\cos \theta + r^2 \cos^2 \theta \iff \\ r^2 (\cos^2 \theta - 1) - 4r \cos \theta + 4 = 0. $$

This is a quadratic equation for $r$ whose solutions are

$$ r = \frac{4 \cos \theta \pm \sqrt{16 \cos^2 \theta - 16(\cos^2 \theta - 1)}}{2 \cos^2 \theta - 2} = \frac{4 (\cos \theta \pm 1)}{2 (\cos^2 \theta - 1)}. $$

Since $r$ is non-negative and the denominator is non-positive, we must take the solution

$$ r = \frac{4(\cos \theta - 1)}{2 (\cos \theta - 1)(\cos \theta + 1)} = \frac{2}{\cos \theta + 1}.$$

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  • $\begingroup$ why not use $x^2+y^2=r^2$? this is unnecessarily complicated. $\endgroup$
    – Dando18
    Aug 13, 2017 at 4:35
  • $\begingroup$ @Dando18: Well, obviously I used this in the second line. I just wanted to start with something that "always" works - plug in the new coordinates and simplify. Of course you can make various shortcuts but formally speaking, even if you don't know how to simplify your expression, the resulting equation is the correct equation in polar coordinates for the curve. $\endgroup$
    – levap
    Aug 13, 2017 at 4:38
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$$ \begin{align} x^2+y^2&=(2-x)^2\\ (r\,\cos\theta)^2+(r\,\sin\theta)^2&=(2-r\,\cos\theta)^2\\ r^2(\cos^2\theta+\sin^2\theta)&=(2-r\,\cos\theta)^2\\ r^2&=(2-r\,\cos\theta)^2\\ r&=2-r\,\cos\theta\\ r+r\,\cos\theta&=2\\ r(1+\cos\theta)&=2\\ r&=\frac{2}{1+\cos\theta}\\ \end{align} $$

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