Circles are all around you in everyday life, from tires on cars to buttons on coats, as well as on the tops of bowls, glasses, and water bottles. Ellipses are less common. One example is the orbits of planets, but you should be able to find the area of a circle or an ellipse, or the circumference of a circle, based on information given to you in a problem. Circles and ellipses are examples of conic sections, which are curves formed by the intersection of a plane with a cone.

Some sample problems are shown below, with solutions worked out.

Exercise 1

Let's say you are a gardener, and you have just planted a lot of flowers that you want to water. The flower bed is 15 feet wide, and 15 feet long. You are using a circular sprinkler system, and the water reaches 6 feet out from the center. The sprinkler is located, from the bottom left corner of the bed, 7 feet up, and 6 feet over.

1. If the flower bed was a graph with the bottom left corner being the origin, what would the equation of the circle be?
2. What is the area being watered by the sprinkler?
3. What percentage of the garden that is being watered?

Answer

If the bottom left corner is the origin, it has coordinates $\left(0,0\right)$. From there the sprinkler is 6 feet over, so the $x$-coordinate of the center is 6. The sprinkler is 7 feet up, so the $y$-coordinate of the center is 7. Ergo, the center of the circle is at coordinate $\left(6,7\right)$. The water reaches 6 feet out from the sprinkler, so the circle radius is 6 feet. Therefore the equation of this circle is:

$\displaystyle{\left(x-6\right)^2+\left(y-7\right)^2=36}$

The area that is watered by the sprinkler can be labeled $A_{sprinkler}$, and is:

$\displaystyle{ \begin{aligned} A_{sprinkler} &=\pi\cdot r^2\\ &=\pi\cdot 6^2\\ &=36\pi \end{aligned} }$

The first step to finding the percentage of the garden that is being watered is to check that none of the water is falling outside the garden. The sprinkler is at coordinate $\left(6,7\right)$, and the radius of the sprinkler is 6 feet. You can test that the water does not fall outside the $15\times 15$ garden. 

Once we know that the area that is watered is completely inside the garden, the percentage of the garden that is watered can be found by dividing the area watered by the total area of the garden, and then multiplying by $100\%$:

$\displaystyle{ \begin{aligned} Percentage_{watered} &=\frac {A_{sprinkler}}{A_{flower bed}}\cdot 100\%\\ &=\frac {36\pi}{15^2}\cdot 100\%\\ &=\frac {113.1}{225}\cdot 100\%\\ &=50.3\% \end{aligned} }$

Exercise 2

Now, let's take it the other way. $(x - 4)^2+(y+8)^2=49$ is the equation for a circle.

1. What are the coordinates of the center of the circle?
2. What is the radius?
3. Draw the circle. 
4. Find two points on the circle and plug them into the equation to make sure your drawing is correct. 

Answer

The center of the circle can be found by comparing the equation in this exercise to the equation of a circle:

$\left(x-h\right)^2+\left(y-k\right)^2=r^2$

The center of the circle is at coordinate $\left(h,k\right)$, and so the center of the circle in this exercise is at $\left(4,-8\right)$.

The radius of the circle is $r$. From the equation of a circle, $\displaystyle{r^2=49}$. Therefore: 

$\displaystyle{ \begin{aligned} r^2 &= 49 \\ r &= \sqrt{49} \\ &= 7 \end{aligned} }$

Exercise 2 circle

Graph of the circle $(x - 4)^2+(y+8)^2=49$.

There are many points you could choose. For example, the point $\left(4, -1\right)$ is at the top of the circle. Plugging this into the equation, we get:

 $\displaystyle{ \begin{aligned} \left(x-h\right)^2+\left(y-k\right)^2 &=r^2 \\ (4-4)^2+(-1+8)^2 &= 49 \\ (0)^2 + (7)^2 &= 49 \\ 49 &= 49 \end{aligned} }$

The left side is equal to the right side of the equation, and so this is a valid point on the circle. 

 The leftmost point on the circle is $(-3,-8)$. Plugging this in to the equation:

$\displaystyle{ \begin{aligned} \left(x-h\right)^2+\left(y-k\right)^2 &=r^2 \\ \left(-3-4\right)^2+\left(-8+8\right)^2 &= 49 \\ (-7)^2 + (0)^2 &= 49 \\ 49 &= 49 \end{aligned} }$

The left side is equal to the right side of the equation, and so this is a valid point on the circle.

Exercise 3

1. Put $2x^2+2y^2+8x+24y+60=0$ into the standard circle form:         $(x-h)^2+(y-k)^2=r^2$.
2. What is the center and the radius of the circle?
3. Draw the circle.
4. Find two points on the circle and plug them into the equation to make sure your drawing is correct.

Answer

First, divide the equation by the coefficient of $x^2$ and $y^2$, which is $2$:

 $\displaystyle{ \begin{aligned} 2x^2+2y^2+8x+24y+60 &=0 \\ \frac{2x^2+2y^2+8x+24y+60}{2} &=\frac{0}{2} \\ \frac{2x^2}{2}+\frac{2y^2}{2}+\frac{8x}{2}+\frac{24y}{2}+\frac{60}{2} &=\frac{0}{2} \\ x^2+y^2+4x+12y+30 &=0 \end{aligned} }$

Next, collect $x$ and $y$ terms together, and bring the number to the right side of the equation:

$(x^2+4x)+(y^2+12y)=-30$

Now, complete the square in both parentheses, subtracting or adding the necessary constant to both sides of the equation:

$\displaystyle{ \begin{aligned} (x^2+4x+4)+(y^2+12y+36) &=-30+4+36 \\ (x^2+4x+4)+(y^2+12y+36) &= 10 \end{aligned} }$

Notice that each term is a perfect square, which gives: 

$(x+2)^2+(y+6)^2=10$

This is now in the standard form for the equation of a circle.

The center of the circle is at coordinate $\left(h,k\right)$, and so the center of the circle in this exercise is at $\left(-2,-6\right)$.

The radius of the circle is $r$. From the equation of a circle, $\displaystyle{r^2=10}$. Therefore: 

$\displaystyle{ \begin{aligned} r^2 &= 10 \\ r &= \sqrt{10} \end{aligned} }$

It is fine to leave the value of $r$ in this form.

Exercise 3 circle

Graph of the circle $2x^2+2y^2+8x+24y+60=0$. 

There are many points you could choose. For example, $(-5,-5)$ and $(-3, -9)$:

$\displaystyle{ \begin{aligned} (-5+2)^2 + (-5 + 6)^2 &= 10 \\ (-3)^2 + (1)^2 &= 10 \\ 9 + 1 &=10\\ 10 &= 10 \end{aligned} }$

$\displaystyle{ \begin{aligned} (-3+2)^2 + (-9+6)^2 &= 10\\ (-1)^2 + (-3)^2 &= 10 \\ 1 + 9 &= 10 \\ 10 &= 10 \end{aligned} }$

Both are valid points on the circle.

Exercise 4

1. Put the equation $\displaystyle{ \frac{4x^2}{9} + 25y^2 = 1 }$ into standard form.
2. What is the center?
3. How long is the major axis?
4. How long is the minor axis?
5. Graph it.

Answer

This almost looks like an ellipse in standard form, doesn't it? It even has a number on the right side. But it isn't. There is no room in the standard form for the values $4$ and $25$ in the numerators. How can we get rid of them to get into standard form?

Rewrite the left-hand term, $\displaystyle{\frac{4x^2}{9}}$, by dividing the top and bottom of the fraction by $4$. Leave the bottom as a fraction; don't make it a decimal.

Rewrite the right-hand term, $25y^2$, as $\displaystyle{\frac{25y^2}{1}}$, and divide the top and bottom of the fraction by $25$. Leave the bottom as a fraction; don't make it a decimal.

The equation is now:

$\displaystyle{ \begin{aligned} \frac{4x^2}{9} + 25y^2 &= 1 \\ \frac{\frac{4x^2}{4}}{\frac{9}{4}} + \frac{\frac{25y^2}{25}}{\frac{1}{25}} &= 1 \\ \frac{x^2}{\frac{9}{4}} + \frac{y^2}{\frac{1}{25}} &= 1 \end{aligned} }$

From the standard equation, $\displaystyle{\frac {\left(x-h\right)^2}{a^2}+\frac {\left(y-k\right)^2}{b^2}= 1}$, we know that the center is at $\left(h,k\right)$. Since these are both zero in this equation, the center is at $\left(0,0\right)$.

The major axis depends which is longer, $a$ or $b$. Let's solve for both, and find out which is larger afterward.

$\displaystyle{ \begin{aligned} a^2 &= \frac{9}{4} \\ a &= \frac{3}{2} \\ 2a &= \frac{6}{2} \\ &= 3 \end{aligned} }$

$\displaystyle{ \begin{aligned} b^2 &= \frac{1}{25} \\ b &= \frac{1}{5} \\ 2b &= \frac{2}{5} \end{aligned} }$ 

Since $\displaystyle{3>\frac{2}{5}}$, $2a>2b$, and $2a$ is the major axis. The major axis length is $2a = 3$, and the minor axis length is $\displaystyle{2b = \frac{2}{5}}$. 

Exercise 4 ellipse

Graph of the ellipse $ \frac{4x^2}{9} + 25y^2 = 1$.