Converting an equation from Cartesian to Polar form?

Question

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.

Answer 1

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 $r$, $$x^2 + y^2 = (2-x)^2$$ $$r^2 = (2-x)^2$$ $$ r = 2-x$$ $$ r = 2 - r\cos\left(\theta\right)$$ $$ r + r\cos\left(\theta\right) = 2$$ $$ r(1 + \cos\left(\theta\right)) = 2$$ $$ r = \frac{2}{1 + \cos\left(\theta\right)}$$

Answer 2

$$x^2 +y^2 =(2-x)^2\implies y^2+4x-4=0$$ 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 $r$.

Answer 3

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

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

Answer 4

$$ \begin{align} x^2+y^2&=(2-x)^2\\ (r\,\cos\theta)^2+(r\,\sin\theta)^2&=(2-r\,\cos\theta)^2\\ r^2(\cos^2\theta+\sin^2\theta)&=(2-r\,\cos\theta)^2\\ r^2&=(2-r\,\cos\theta)^2\\ r&=2-r\,\cos\theta\\ r+r\,\cos\theta&=2\\ r(1+\cos\theta)&=2\\ r&=\frac{2}{1+\cos\theta}\\ \end{align} $$