# Question 5b70d

Mar 7, 2017

$r = \pm 1$

#### Explanation:

This is the unit circle about the Origin

• Take the parameterisation in polar coordinates:

$x = r \cos \theta , y = r \sin \theta$

• Observe that:

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

• Then note that you actually have:

${r}^{2} = 1 , r = \pm 1$

I'd go with the $\pm$ bit too. I've seen it argued that $r$ should only ever be positive, though I can't remember the detail and it makes no sense to me (albeit I say that as a user of maths as opposed to mathematician).

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Mar 7, 2017

${r}^{2} = 1$

#### Explanation:

To convert from $\textcolor{b l u e}{\text{cartesian to polar form}}$

color(red)(bar(ul(|color(white)(2/2)color(black)(x=rcostheta;y=rsintheta)color(white)(2/2)|)))#

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

$\Rightarrow {\left(r \cos \theta\right)}^{2} + {\left(r \sin \theta\right)}^{2} = 1$

$\Rightarrow {r}^{2} {\cos}^{2} \theta + {r}^{2} {\sin}^{2} \theta = 1$

$\Rightarrow {r}^{2} \left({\cos}^{2} \theta + {\sin}^{2} \theta\right) = 1$

$\Rightarrow {r}^{2} = 1 \text{ since } {\cos}^{2} \theta + {\sin}^{2} \theta = 1$

This is the equation of a circle, centred at the origin with radius 1