$\text { Find the polar coordinates, } 0 \leq \theta

Question

$\text { Find the polar coordinates, } 0 \leq \theta < 2 \pi \text { and } r \geq 0 \text {, of the point given in Cartesian coordinates. }$
-(-2, 0) 
A) $( - 2 , \pi )$
B) $\left( 2 , \frac { 3 \pi } { 2 } \right)$
C) $( 2 , \pi )$
D) $\left( 2 , \frac { \pi } { 2 } \right)$

Quizplus · Accepted Answer

We use the formulas 
$$ x = r \cos \theta \quad \text {and} \quad y = r \sin \theta $$
to find the polar coordinates. Since the point is (-2, 0), we have 
$$ r \cos \theta = -2 \quad \text {and} \quad r \sin \theta = 0. $$
Solving the first equation for $\cos \theta$, we get $\cos \theta = -\frac{2}{r}$. Since $0 \leq \theta < 2\pi$, we must have $\cos \theta \leq 0$, which means $r$ must be positive and $\theta$ must be in the second or third quadrant. Solving the second equation for $\sin \theta$, we get $\sin \theta = 0$, which means $\theta$ is an integer multiple of $\pi$. Putting it all together, we have
$$ \cos \theta = -\frac{2}{r} \quad \text {and} \quad \theta = \pi k, $$
where $r > 0$ and $k$ is an integer. To determine $r$, we use the first equation:
$$ r \cos \theta = -2 \implies r \cdot \frac{-2}{r} = -2 \implies r = 2. $$
(Note that multiplying both sides by $r$ is only valid when $r \neq 0$, but we already ruled out $r=0$ because $(-2,0)$ is not the origin.)

Therefore, we have $r=2$ and $\cos \theta = -\frac{2}{2} = -1$, which means $\theta = \pi$. Since $0 \leq \theta < 2\pi$, we choose $\theta = \pi$ and the polar coordinates are $(2,\pi)$.