# Consider the circle C of radius 6, centered at the origin. a) Find a parametrization for C...

## Question:

Consider the circle {eq}C {/eq} of radius {eq}6 {/eq}, centered at the origin.

a) Find a parametrization for {eq}C {/eq} inducing a counterclockwise orientation and starting at {eq}(6, 0). {/eq}

b) Find a parametrization for {eq}C {/eq} inducing a clockwise orientation and starting at {eq}(0, 6). {/eq}

c) Find a parametrization for {eq}C {/eq} if it is now centered at the point {eq}(2, 7). {/eq}

## Parametric Equations

A circle {eq}(x-h)^2+(y-k)^2=r^2 {/eq} in Cartesian (rectangular) coordinates can be defined parametrically with a function {eq}\textbf{r}:\left[0,2\pi\right]\rightarrow\mathbb{R}^2 {/eq}. As {eq}t {/eq} increases from {eq}t=0 {/eq} to {eq}t=2\pi {/eq} the circle is either traversed in a clockwise or counter clockwise direction. For example, suppose the circle has center {eq}(h,k) {/eq} and radius {eq}r {/eq} and is traversed counter clockwise. In this case, the parametric function defined as

{eq}\begin{eqnarray*} \textbf{r}(t) &=& (h+r\cos(t),k+r\sin(t)) \end{eqnarray*} {/eq}

would be the correct parametrization. Parts a), b) and c) explore different ways a circle can be parametrized.

## Answer and Explanation:

The question is restated. Consider the circle {eq}C {/eq} of radius 6 centered at the origin.

a) Find a parametrization for {eq}C {/eq} inducing a counter clockwise orientation and starting at the point {eq}(6,0) {/eq}.

b) Find a parametrization for {eq}C {/eq} inducing a clockwise orientation and starting at {eq}(0,6) {/eq}.

c) Find a parametrization for {eq}C {/eq} if it is now centered at the point {eq}(2,7) {/eq}.

Consider part a). The equation of the circle in Cartesian (rectangular) coordinates is

{eq}\begin{eqnarray*} (x-0)^2+(y-0)^2 &=& 6^2 \\ \frac{(x-0)^2}{6^2}+\frac{(y-0)^2}{6^2} &=& 1 \\ \left(\frac{x-0}{6}\right)^2+\left(\frac{y-0}{6}\right)^2 &=& 1. \end{eqnarray*} {/eq}

Consider the trigonometric identity

{eq}\begin{eqnarray*} \cos^2(t)+\sin^2(t) &=& 1. \end{eqnarray*} {/eq}

Comparing the trigonometric identity with the equation of the circle we see that

{eq}\begin{eqnarray*} \frac{x-0}{6} &=& \cos(t) \\ x &=& 0+6\cos(t) \\ \frac{y-0}{6} &=& \sin(t) \\ y &=& 0+6\sin(t) \end{eqnarray*} {/eq}

is a natural way to define the circle parametrically. If {eq}t=0 {/eq} the corresponding point is {eq}x=6\cos(0)=6 {/eq} and {eq}y=6\sin(0)=0 {/eq}, that is, the point {eq}(6,0) {/eq}. If {eq}t=\frac{\pi}{2} {/eq}, for example,

{eq}\begin{eqnarray*} x &=& 6\cos\left(\frac{\pi}{2}\right) \\ &=& 0 \\ y &=& 6\sin\left(\frac{\pi}{2}\right) \\ &=& 6. \end{eqnarray*} {/eq}

That is, the point is {eq}(0,6) {/eq}. So, as {eq}t {/eq} goes from {eq}t=0 {/eq} to {eq}t=\frac{\pi}{2} {/eq} the direction around the circle is counter clockwise. The circle is completely traversed when {eq}t=2\pi {/eq}. Therefore, the parametrization for {eq}C {/eq} inducing a counter clockwise orientation and starting at the point {eq}(6,0) {/eq} is {eq}\textbf{r}:\left[0,2\pi\right]\rightarrow\mathbb{R}^2 {/eq} defined as

{eq}\begin{eqnarray*} \textbf{r}(t) &=& (6\cos(t),6\sin(t)). \end{eqnarray*} {/eq}

Consider part b). The point that corresponds to {eq}t=0 {/eq} is {eq}(0,6) {/eq}. Moreover, the orientation is clockwise. The parametrization for {eq}C {/eq} inducing a clockwise orientation and starting at the point {eq}(0,6) {/eq} is {eq}\textbf{r}:\left[0,2\pi\right]\rightarrow\mathbb{R}^2 {/eq} defined as

{eq}\begin{eqnarray*} \textbf{r}(t) &=& \left(6\cos\left(\frac{\pi}{2}-t\right),6\sin\left(\frac{\pi}{2}-t\right)\right). \end{eqnarray*} {/eq}

The point corresponding to {eq}t=0 {/eq} is

{eq}\begin{eqnarray*} \textbf{r}(0) &=& \left(6\cos\left(\frac{\pi}{2}-0\right),6\sin\left(\frac{\pi}{2}-0\right)\right) \\ &=& \left(6\cos\left(\frac{\pi}{2}\right),6\sin\left(\frac{\pi}{2}\right)\right) \\ &=& (0,6). \end{eqnarray*} {/eq}

The point corresponding to {eq}t=\frac{\pi}{2} {/eq} is

{eq}\begin{eqnarray*} \textbf{r}\left(\frac{\pi}{2}\right) &=& \left(6\cos\left(\frac{\pi}{2}-\frac{\pi}{2}\right),6\sin\left(\frac{\pi}{2}-\frac{\pi}{2}\right)\right) \\ &=& (6\cos(0),6\sin(0)) \\ &=& (6,0). \end{eqnarray*} {/eq}

The point corresponding to {eq}t=\pi {/eq} is

{eq}\begin{eqnarray*} \textbf{r}(0) &=& \left(6\cos\left(\frac{\pi}{2}-\pi\right),6\sin\left(\frac{\pi}{2}-\pi\right)\right) \\ &=& \left(6\cos\left(-\frac{\pi}{2}\right),6\sin\left(-\frac{\pi}{2}\right)\right) \\ &=& (0,-6). \end{eqnarray*} {/eq}

It is clear from this analysis that the circle starts at {eq}(0,6) {/eq} goes to {eq}(6,0) {/eq} and then to {eq}(0,-6) {/eq} in a clockwise direction. Hence, this is the desired parametrization.

Consider part c). The orientation was not specified in the question. A counter clockwise orientation will be used. Since the center of the circle is {eq}(2,7) {/eq} then the equations for {eq}x {/eq} and {eq}y {/eq} should be

{eq}\begin{eqnarray*} \frac{x-2}{6} &=& \cos(t) \\ x &=& 2+6\cos(t) \\ \frac{y-7}{6} &=& \sin(t) \\ y &=& 7+6\sin(t). \end{eqnarray*} {/eq}

The parametrization for {eq}C {/eq} with a center {eq}(2,7) {/eq}, radius 6 and inducing a counter clockwise orientation is {eq}\textbf{r}:\left[0,2\pi\right]\rightarrow\mathbb{R}^2 {/eq} defined as

{eq}\begin{eqnarray*} \textbf{r}(t) &=& (2+6\cos(t),7+6\sin(t)). \end{eqnarray*} {/eq}

Note that the circle is traversed starting at the point {eq}(8,7) {/eq} when {eq}t=0 {/eq} and then going to {eq}(2,13) {/eq} when {eq}t=\frac{\pi}{2} {/eq}.

#### Learn more about this topic:

from Precalculus: High School

Chapter 24 / Lesson 3