# Question #65a2a

Apr 16, 2017

${x}^{2} + {y}^{2} = \sqrt{{x}^{2} + {y}^{2}} + 2 x$

intersection coordinates are $\left(0 , - 1\right)$ and $\left(0 , 1\right)$

#### Explanation:

Our Polar equation is:

$r = 1 + 2 \cos \theta$

To convert from polar to rectangular (Cartesian) coordinates we use the fundamental relationship:

$\left.\begin{matrix}x = r \cos \theta \\ y = r \sin \theta\end{matrix}\right. \setminus \setminus$ and $\setminus \setminus {x}^{2} + {y}^{2} = {r}^{2}$

So we can transform our equation as follows:

$r = 1 + 2 \cos \theta$
$\therefore {r}^{2} = r + 2 r \cos \theta$
$\therefore {x}^{2} + {y}^{2} = \sqrt{{x}^{2} + {y}^{2}} + 2 x$

For the interaction with a unit circle which would have the equation:

${x}^{2} + {y}^{2} = 1$

We would have:

$\therefore 1 = 1 + 2 x \implies 2 x = 0$
$\text{ } \implies x = 0$

And:

$x = 0 \implies {y}^{2} = 1$
$\text{ } \implies y = \pm 1$

So the intersection coordinates would be $\left(0 , - 1\right)$ and $\left(0 , 1\right)$