Find the point(s) of intersection of the curves $r = 2$ and $r = 4 \cos \theta$.
A) $\left( 2 , \frac { \pi } { 6 } \right) , \left( 2 , - \frac { \pi } { 6 } \right)$
B) $\left( 2 , \frac { \pi } { 4 } \right) , \left( 2 , - \frac { \pi } { 4 } \right)$
C) $\left( 2 , \frac { \pi } { 3 } \right) , \left( 2 , - \frac { \pi } { 3 } \right)$
D) $\left( 2 , \frac { \pi } { 6 } \right)$
E) $\left( 2 , \frac { \pi } { 3 } \right)$

Question

Quizplus · Accepted Answer

To find the intersection, set $r = 2$ equal to $r = 4 \cos \theta$. Solving $2 = 4 \cos \theta$ gives $\cos \theta = \frac{1}{2}$, which corresponds to $\theta = \frac{\pi}{3}$ and $\theta = -\frac{\pi}{3}$. Thus, the points of intersection are $\left( 2 , \frac { \pi } { 3 } \right)$ and $\left( 2 , - \frac { \pi } { 3 } \right)$.

Find the Point(s) of Intersection of the Curves r=2r = 2r=2

Find the Point(s) of Intersection of the Curves $r = 2$