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# Parametrize a circle – Equations, Graphs, and Examples

Learning how we can parametrize a circle is helpful, especially when we want to visualize a given objectâ€™s position over time. As with other applications of parametric equations, it can help us model relationships that do not necessarily function themselves.

* We can parametrize a circle by expressing *$\boldsymbol{x}$

*$\boldsymbol{x}$*

**and**

**in terms of cosine and sine, respectively.**Weâ€™ve already learned about parametric equations in the past, and this article is an extension of that knowledge â€“ focusing on the process of parametrizing circles. Before we dive right into the essential topic for this article, make sure to review your knowledge on the following:

Review what the general equation for the circle is as a conic section.

Understand how the trigonometric ratios can be obtained on a Cartesian coordinate system.

Ready to learn more? Letâ€™s go ahead and begin by reviewing what we know of parametric equations and see how we can apply these to circles.

**How to parametrize a circle?Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â **

When given an equation in rectangular form, we can express $x$ and $y$ as a function of $t$. The new element, $t$, is now our new parameter, hence, the name of the relationship shared by $x$, $y$, and $t$.

\begin{aligned}x &= f(t)\\y&= g(t)\end{aligned}

This means that we can rewrite the equation of the circle, $x^2 + y^2 = r^2$, in terms of $t$. We can do this by assigning a special function for $x$ and $y$ – this is where the unit circle and trigonometric ratios enter.

Letâ€™s begin by parametrizing the unit circleâ€™s equation,$x^2 + y^2 = 1$, before generalizing the rules.

**Parametrizing a Unit Circle**

Recall that when $0 \leq t \leq 2\pi$, we can use the expression $x$ and $y$ in terms of cosine and sine.

\begin{aligned}x &= \cos t\\y &= \sin t\end{aligned}

If we square both sides of the equation and add the two, weâ€™ll develop the unit circle’s parametric form.

\begin{aligned}x^2 &= \cos^2 t\\y &= \sin^2 t\\\\x^2 + y^2 &=1\\\cos^2 t + \sin^2 t &= 1\end{aligned}

**Parametrizing a Circle Centered at **$\boldsymbol{(0, 0)}$

We can extend this concept with circles centered at the origin but with a radius $r$.This means that $r$ will now be a factor of the parametric forms of $x$ and $y$, as shown below.

\begin{aligned}x&= r\cos t\\y&= r\sin t\end{aligned}

Squaring both equations, weâ€™ll be able to come up with the parametric form of the circleâ€™s equation.

\begin{aligned}x^2 &= r^2\cos^2 t\\y &= r^2\sin^2 t\\x^2 + y^2 &= r^2\\r^2(\cos^2 t + \sin^2 t) &= r^2\end{aligned}

The third case weâ€™ll have to consider is parametrizing a circle that is not centered at the origin.

**Parametrizing a Circle Centered at **$\boldsymbol{(h, k)}$

Since the $x$ and $y$ coordinates are translated along $h$ and $k$ units, respectively, weâ€™ll have to consider these for the parametrized form of the circleâ€™s equation.

\begin{aligned}x -h&= r\cos t\\x &= h+ r\cos t\\y – k&= r\sin t\\y &= k + r\sin t\end{aligned}

Letâ€™s square both sides of the equation and observe how the equation’s parametric would look.

\begin{aligned}(x- h)^2 &= r^2\cos^2 t\\(y -k)^2 &= r^2\sin^2 t\\\\(x- h)^2 + (y- k)^2 &= r^2\\r^2(\cos^2 t + \sin^2 t) &= r^2\end{aligned}

Now that we know how we can parametrize a circleâ€™s equation, why donâ€™t we see how this affects the graph of the equations?

**Graphing the p****arametric equations of a circle**

Parametric equations are best used when we want to account for the direction of the curve. These are some quick pointers to remember when graphing the parametric curve of a circle.

Take note of the assigned interval for $t$.

Assign key values of $t$ and use these values to find the corresponding values of $x$ and $y$.

Plot the parametric curve and account for the circle’s direction based on the resulting table of values.

Why donâ€™t we try graphing the parametric equations shown below and the assigned intervals for $t$.

\begin{aligned}x &= 2\cos t\\y&= 2\sin t\\0&\leq t\leq 2\pi\end{aligned}

We begin by assigning key values for $t$ between the interval, $0\leq t\leq 2\pi$, such as $t = \left\{0, \dfrac{\pi}{2}, \pi, \dfrac{3\pi}{2}\right\}$.

$\boldsymbol{t}$ | $\boldsymbol{x = 2\cos t}$ | $\boldsymbol{y = 2\sin t}$ |

$0$ | \begin{aligned}x &= 2\cos 0\\&=2 \end{aligned} | \begin{aligned}y &= 2\sin 0\\&=0 \end{aligned} |

$\dfrac{\pi}{2}$ | \begin{aligned}x &= 2\cos \dfrac{\pi}{2}\\&=0 \end{aligned} | \begin{aligned}y &= 2\sin \dfrac{\pi}{2}\\&= 2 \end{aligned} |

$\pi$ | \begin{aligned}x &= 2\cos \pi\\&=-2 \end{aligned} | \begin{aligned}y &= 2\sin \pi\\&=0 \end{aligned} |

$\dfrac{3\pi}{2}$ | \begin{aligned}x &= 2\cos \dfrac{3\pi}{2}\\&=0 \end{aligned} | \begin{aligned}y &= 2\sin \dfrac{3\pi}{2}\\&= -2 \end{aligned} |

This means that the arrows will be in counterclockwise direction starting from $(2, 0)$ to $(-2, 0)$ then back to $(2, 0)$. We can now graph the parametric equations and letâ€™s not forget to include the arrows to reflect the direction of the curve.

This graph shows the importance of parametric equations â€“ we can graph the motions of objects while accounting for a third parameter – $t$. Weâ€™ll apply similar approaches when graphing other circles using their parametric equations.

**Example 1**

Write two sets of parametric equations for the following rectangular equations.

Â Â a. $x^2 + y^2 = 64$

Â Â b. $x^2 + y^2 = 12$

Â Â c. $(x â€“ 3)^2 + (y + 2)^2 = 81$

Â Â d. $(x + 6)^2 + (y – 4)^2 = 24$

__Solution__

The equation, $x^2 + y^2 = 64$, is a circle centered at the origin, so the standard form the parametric equations representing the curve will be \begin{aligned}x &=r\cos t\\y &=r\sin t\\0&\leq t\leq 2\pi\end{aligned}, where $r$ represents the radius of the circle.

Since $r^2 = 64$, the value of the circleâ€™s radius is equal to $r= 8$. Hence, we have the parametric equations shown below. Weâ€™ve also included the interval for $t$.

\begin{aligned}x &=8\cos t\\y &=8\sin t\\0&\leq t \leq 2\pi\end{aligned}

Weâ€™ll apply a similar process in writing the parametric equations of $x^2 + y^2 = 12$, but this time, we have $r^2 = 12$ and consequently, $r = 2\sqrt{3}$.

\begin{aligned}x &=2\sqrt{3}\cos t\\y &=2\sqrt{3}\sin t\\0 &\leq t\leq 2\pi\end{aligned}

For the third equation,$(x â€“ 3)^2 + (y + 2)^2 = 81$, the circle is now centered at $(3, -2)$, so we must also translate the parametric equations accordingly.

Weâ€™ll use the equations, $\left\{\begin{matrix}x =h + r\cos t\\y =k + r\sin t\end{matrix}\right.$, this time, where $(h, k)$ represents the circleâ€™s center. For our rectangular equation,we also have $r^2 = 81$, so $r = 9$.

\begin{aligned}x &=3 + 9\cos t\\y &=-2 + 9\sin t\\0&\leq t\leq 2\pi\end{aligned}

The fourth equation,$(x + 6)^2 + (y – 4)^2 = 12$, looks similar to the third one, so weâ€™ll use $(h,k) = (-6, 4)$. Since $r^2 = 24$, we can use $r =2\sqrt{6}$.

\begin{aligned}x &=-6 + 2\sqrt{6}\cos t\\y &=4 + 2\sqrt{6}\sin t\\0&\leq t\leq 2\pi\end{aligned}

Hence, weâ€™ve shown how we can write an equation of a circle into its parametric form.

**Example ****2**

Write two sets of parametric equations for the following rectangular equations. Use the resulting parametric equations to graph the circle (weâ€™ll assume that $0 \leq t\leq 2\pi$).

Â Â a. $x^2 + y^2 = 36$

Â Â b. $(x + 3)^2 + (y – 1)^2 = 16$

__Solution__

Since the first rectangular equation shows a circle centered at the origin, the standard form of the parametric equations are$\left\{\begin{matrix}x =r\cos t\\y =r\sin t\\0\leq t\leq 2\pi\end{matrix}\right.$. We have $r^2 = 36$, so $r = 6$.

Hence, the circleâ€™s parametric equations are as shown below.

\begin{aligned}x &=6\cos t\\y &=6\sin t\\0&\leq t\leq 2\pi\end{aligned}

To graph this set of parametric equations, letâ€™s use key values of $t$ as guides to know how the parametric curve behaves.

$\boldsymbol{t}$ | $\boldsymbol{x = 6\cos t}$ | $\boldsymbol{y = 6\sin t}$ |

$0$ | \begin{aligned}x &= 6\cos 0\\&=6 \end{aligned} | \begin{aligned}y &= 6\sin 0\\&=0 \end{aligned} |

$\dfrac{\pi}{2}$ | \begin{aligned}x &= 6\cos \dfrac{\pi}{2}\\&=0 \end{aligned} | \begin{aligned}y &= 6\sin \dfrac{\pi}{2}\\&= 6 \end{aligned} |

$\pi$ | \begin{aligned}x &= 6\cos \pi\\&=-6 \end{aligned} | \begin{aligned}y &= 6\sin \pi\\&=0 \end{aligned} |

$\dfrac{3\pi}{2}$ | \begin{aligned}x &= 6\cos \dfrac{3\pi}{2}\\&=0 \end{aligned} | \begin{aligned}y &= 6\sin \dfrac{3\pi}{2}\\&= -6 \end{aligned} |

This means that the parametric curveâ€™s direction is counterclockwise, from $(6,0)$ when $t = 0$, $(0,-6)$ when $t = \dfrac{3\pi}{2}$, then back to $(6,0)$ when $t = 2\pi$.

Now, for the second equation,$(x + 3)^2 + (y – 1)^2 = 16$ , we can see that the circle has a radius of $4$ and is centered at $(-3, 1)$**. **Weâ€™ll use these components to rewrite the equation in their parametric forms.

\begin{aligned}x &=h + r\cos t\\y &=k + r\sin t\\\\x &=-3 + 4\cos t\\y &=1 + 4\sin t\phantom{x}\end{aligned}

To graph the parametric curve, weâ€™ll once again use the same interval, $0\leq t\leq 2\pi$, and select key values for $t$ as shown in the table of values below.

Â | $\boldsymbol{x = -3 + 4\cos t}$ | $\boldsymbol{y =1 + 4\sin t}$ |

$0$ | \begin{aligned}x &= -3 +4\cos 0\\&= -3+4\\&= 1 \end{aligned} | \begin{aligned}y &= 1 +4\sin 0\\&=1 +0\\&=1 \end{aligned} |

$\dfrac{\pi}{2}$ | \begin{aligned}x &= -3 +4\cos \dfrac{\pi}{2}\\&=-3 +0\\&= -3 \end{aligned} | \begin{aligned}y &= 1+ 4\sin \dfrac{\pi}{2}\\&= 1+4\\&=5 \end{aligned} |

$\pi$ | \begin{aligned}x &= -3 +4\cos \pi\\&=-3- 4\\&= -7 \end{aligned} | \begin{aligned}y &= 1+4\sin \pi\\&=1 + 0\\&=1\end{aligned} |

$\dfrac{3\pi}{2}$ | \begin{aligned}x &= -3 +4\cos \dfrac{3\pi}{2}\\&=-3+0\\&= -3 \end{aligned} | \begin{aligned}y &= 1+ 4\sin \dfrac{3\pi}{2}\\&= 1-4\\&= -3 \end{aligned} |

This means that the parametric curve will pass through these points in this order:$(1,1)\rightarrow (-3, 5)\rightarrow (-7,1) \rightarrow (-3, -3)\rightarrow (1, 1)$, in counterclockwise direction.

### Practice Questions

### Open Problems

From the last two problems, use the resulting parametric equations to graph the circle (weâ€™ll assume that $0 \leq t\leq 2\pi$).

1. $x^2 + y^2 = 9$

2. $(x + 1)^2 + (y – 4)^2 = 36$

### Open Problem Solutions

1.Â

2.Â

*Images/mathematical drawings are created with GeoGebra.*