Learning Objectives
 Use the Distance Formula
 Use the Midpoint Formula
 Write the equation of a circle in standard form
 Graph a circle
Be Prepared 11.1
Before you get started, take this readiness quiz.
 Find the length of the hypotenuse of a right triangle whose legs are 12 and 16 inches.
If you missed this problem, review Example 2.34.  Factor: ${x}^{2}18x+81.$
If you missed this problem, review Example 6.24.  Solve by completing the square: ${x}^{2}12x12=0.$
If you missed this problem, review Example 9.22.
In this chapter we will be looking at the conic sections, usually called the conics, and their properties. The conics are curves that result from a plane intersecting a double cone—two cones placed pointtopoint. Each half of a double cone is called a nappe.
There are four conics—the circle, parabola, ellipse, and hyperbola. The next figure shows how the plane intersecting the double cone results in each curve.
Each of the curves has many applications that affect your daily life, from your cell phone to acoustics and navigation systems. In this section we will look at the properties of a circle.
Use the Distance Formula
We have used the Pythagorean Theorem to find the lengths of the sides of a right triangle. Here we will use this theorem again to find distances on the rectangular coordinate system. By finding distance on the rectangular coordinate system, we can make a connection between the geometry of a conic and algebra—which opens up a world of opportunities for application.
Our first step is to develop a formula to find distances between points on the rectangular coordinate system. We will plot the points and create a right triangle much as we did when we found slope in Graphs and Functions. We then take it one step further and use the Pythagorean Theorem to find the length of the hypotenuse of the triangle—which is the distance between the points.
Example 11.1
Use the rectangular coordinate system to find the distance between the points $\left(6,4\right)$ and $\left(2,1\right).$
Solution
Plot the two points. Connect the two points with a line. Draw a right triangle as if you were going to find slope. 

Find the length of each leg.  
Use the Pythagorean Theorem to find d, the distance between the two points. 
${a}^{2}+{b}^{2}={c}^{2}$ 
Substitute in the values.  ${3}^{2}+{4}^{2}={d}^{2}$ 
Simplify.  $\phantom{\rule{0.3em}{0ex}}9+16={d}^{2}$ 
$\phantom{\rule{2em}{0ex}}25={d}^{2}$  
Use the Square Root Property.  $\phantom{\rule{1em}{0ex}}d=5\phantom{\rule{2em}{0ex}}\overline{)d=\mathrm{5}}$ 
Since distance, d is positive, we can eliminate $d=\mathrm{5}.$ 
The distance between the points $\left(6,4\right)$ and $\left(2,1\right)$ is 5. 
Try It 11.1
Use the rectangular coordinate system to find the distance between the points $\left(6,1\right)$ and $\left(2,\mathrm{2}\right).$
Try It 11.2
Use the rectangular coordinate system to find the distance between the points $\left(5,3\right)$ and $\left(\mathrm{3},\mathrm{3}\right).$
The method we used in the last example leads us to the formula to find the distance between the two points $\left({x}_{1},{y}_{1}\right)$ and $\left({x}_{2},{y}_{2}\right).$
When we found the length of the horizontal leg we subtracted $62$ which is ${x}_{2}{x}_{1}.$
When we found the length of the vertical leg we subtracted $41$ which is ${y}_{2}{y}_{1}.$
If the triangle had been in a different position, we may have subtracted ${x}_{1}{x}_{2}$ or ${y}_{1}{y}_{2}.$ The expressions ${x}_{2}{x}_{1}$ and ${x}_{1}{x}_{2}$ vary only in the sign of the resulting number. To get the positive valuesince distance is positive we can use absolute value. So to generalize we will say $\left{x}_{2}{x}_{1}\right$ and $\left{y}_{2}{y}_{1}\right.$
In the Pythagorean Theorem, we substitute the general expressions $\left{x}_{2}{x}_{1}\right$ and $\left{y}_{2}{y}_{1}\right$ rather than the numbers.
$\begin{array}{cccccc}& & & \hfill {a}^{2}& +\hfill & {b}^{2}={c}^{2}\hfill \\ \text{Substitute in the values.}\hfill & & & \hfill {\left(\left{x}_{2}{x}_{1}\right\right)}^{2}& +\hfill & {\left(\left{y}_{2}{y}_{1}\right\right)}^{2}={d}^{2}\hfill \\ \begin{array}{c}\text{Squaring the expressions makes them}\hfill \\ \text{positive, so we eliminate the absolute value}\hfill \\ \text{bars.}\hfill \end{array}\hfill & & & \hfill {\left({x}_{2}{x}_{1}\right)}^{2}& +\hfill & {\left({y}_{2}{y}_{1}\right)}^{2}={d}^{2}\hfill \\ \text{Use the Square Root Property.}\hfill & & & \hfill d& =\hfill & \pm \sqrt{{\left({x}_{2}{x}_{1}\right)}^{2}+{\left({y}_{2}{y}_{1}\right)}^{2}}\hfill \\ \begin{array}{c}\text{Distance is positive, so eliminate the negative}\hfill \\ \text{value.}\hfill \end{array}\hfill & & & \hfill d& =\hfill & \sqrt{{\left({x}_{2}{x}_{1}\right)}^{2}+{\left({y}_{2}{y}_{1}\right)}^{2}}\hfill \end{array}$
This is the Distance Formula we use to find the distance d between the two points $\left({x}_{1},{y}_{1}\right)$ and $\left({x}_{2},{y}_{2}\right).$
Distance Formula
The distance d between the two points $\left({x}_{1},{y}_{1}\right)$ and $\left({x}_{2},{y}_{2}\right)$ is
Example 11.2
Use the Distance Formula to find the distance between the points $\left(\mathrm{5},\mathrm{3}\right)$ and $\left(7,2\right).$
Solution
$\begin{array}{cccccccc}\text{Write the Distance Formula.}\hfill & & & & & d\hfill & =\hfill & \sqrt{{\left({x}_{2}{x}_{1}\right)}^{2}+{\left({y}_{2}{y}_{1}\right)}^{2}}\hfill \\ \text{Label the points,}\phantom{\rule{0.2em}{0ex}}\left(\stackrel{{x}_{1},{y}_{1}}{\mathrm{5},\mathrm{3}}\right),\left(\stackrel{{x}_{2},{y}_{2}}{7,2}\right)\phantom{\rule{0.2em}{0ex}}\text{and substitute.}\hfill & & & & & d\hfill & =\hfill & \sqrt{{\left(7\left(\mathrm{5}\right)\right)}^{2}+{\left(2\left(\mathrm{3}\right)\right)}^{2}}\hfill \\ \\ \text{Simplify.}\hfill & & & & & d\hfill & =\hfill & \sqrt{{12}^{2}+{5}^{2}}\hfill \\ & & & & & d\hfill & =\hfill & \sqrt{144+25}\hfill \\ & & & & & d\hfill & =\hfill & \sqrt{169}\hfill \\ & & & & & d\hfill & =\hfill & 13\hfill \end{array}$
Try It 11.3
Use the Distance Formula to find the distance between the points $\left(\mathrm{4},\mathrm{5}\right)$ and $\left(5,7\right).$
Try It 11.4
Use the Distance Formula to find the distance between the points $\left(\mathrm{2},\mathrm{5}\right)$ and $\left(\mathrm{14},\mathrm{10}\right).$
Example 11.3
Use the Distance Formula to find the distance between the points $\left(10,\mathrm{4}\right)$ and $\left(\mathrm{1},5\right).$ Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.
Solution
$\begin{array}{cccccc}\text{Write the Distance Formula.}\hfill & & & & & d=\sqrt{{\left({x}_{2}{x}_{1}\right)}^{2}+{\left({y}_{2}{y}_{1}\right)}^{2}}\hfill \\ \\ \\ \text{Label the points,}\phantom{\rule{0.2em}{0ex}}\left(\stackrel{{x}_{1},{y}_{1}}{10,\mathrm{4}}\right),\left(\stackrel{{x}_{2},{y}_{2}}{\mathrm{1},5}\right)\phantom{\rule{0.2em}{0ex}}\text{and substitute.}\hfill & & & & & d=\sqrt{{\left(\mathrm{1}10\right)}^{2}+{\left(5\left(\mathrm{4}\right)\right)}^{2}}\hfill \\ \\ \\ \text{Simplify.}\hfill & & & & & d=\sqrt{{\left(\mathrm{11}\right)}^{2}+{9}^{2}}\hfill \\ & & & & & d=\sqrt{121+81}\hfill \\ & & & & & d=\sqrt{202}\hfill \\ \begin{array}{c}\text{Since 202 is not a perfect square, we can leave}\hfill \\ \text{the answer in exact form or find a decimal}\hfill \\ \text{approximation.}\hfill \end{array}\hfill & & & & & \begin{array}{c}d=\sqrt{202}\hfill \\ \phantom{\rule{1em}{0ex}}\text{or}\hfill \\ d\approx 14.2\hfill \end{array}\hfill \end{array}$
Try It 11.5
Use the Distance Formula to find the distance between the points $\left(\mathrm{4},\mathrm{5}\right)$ and $\left(3,4\right).$ Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.
Try It 11.6
Use the Distance Formula to find the distance between the points $\left(\mathrm{2},\mathrm{5}\right)$ and $\left(\mathrm{3},\mathrm{4}\right).$ Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.
Use the Midpoint Formula
It is often useful to be able to find the midpoint of a segment. For example, if you have the endpoints of the diameter of a circle, you may want to find the center of the circle which is the midpoint of the diameter. To find the midpoint of a line segment, we find the average of the xcoordinates and the average of the ycoordinates of the endpoints.
Midpoint Formula
The midpoint of the line segment whose endpoints are the two points $\left({x}_{1},{y}_{1}\right)$ and $\left({x}_{2},{y}_{2}\right)$ is
To find the midpoint of a line segment, we find the average of the xcoordinates and the average of the ycoordinates of the endpoints.
Example 11.4
Use the Midpoint Formula to find the midpoint of the line segments whose endpoints are $\left(\mathrm{5},\mathrm{4}\right)$ and $\left(7,2\right).$ Plot the endpoints and the midpoint on a rectangular coordinate system.
Solution
Write the Midpoint Formula.  $\left(\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1}+{y}_{2}}{2}\right)$ 
Label the points, $\left(\stackrel{{x}_{1},{y}_{1}}{\mathrm{5},\mathrm{4}}\right),\left(\stackrel{{x}_{2},{y}_{2}}{7,2}\right)$ and substitute. 
$\left(\frac{\mathrm{5}+7}{2},\frac{\mathrm{4}+2}{2}\right)$ 
Simplify.  $\left(\frac{2}{2},\frac{\mathrm{2}}{2}\right)$ 
$\left(1,\mathrm{1}\right)$ The midpoint of the segment is the point $\left(1,\mathrm{1}\right).$ 

Plot the endpoints and midpoint. 
Try It 11.7
Use the Midpoint Formula to find the midpoint of the line segments whose endpoints are $\left(\mathrm{3},\mathrm{5}\right)$ and $\left(5,7\right).$ Plot the endpoints and the midpoint on a rectangular coordinate system.
Try It 11.8
Use the Midpoint Formula to find the midpoint of the line segments whose endpoints are $\left(\mathrm{2},\mathrm{5}\right)$ and $\left(6,\mathrm{1}\right).$ Plot the endpoints and the midpoint on a rectangular coordinate system.
Both the Distance Formula and the Midpoint Formula depend on two points, $\left({x}_{1},{y}_{1}\right)$ and $\left({x}_{2},{y}_{2}\right).$ It is easy to confuse which formula requires addition and which subtraction of the coordinates. If we remember where the formulas come from, is may be easier to remember the formulas.
Write the Equation of a Circle in Standard Form
As we mentioned, our goal is to connect the geometry of a conic with algebra. By using the coordinate plane, we are able to do this easily.
We define a circle as all points in a plane that are a fixed distance from a given point in the plane. The given point is called the center, $\left(h,k\right),$ and the fixed distance is called the radius, r, of the circle.
Circle
A circle is all points in a plane that are a fixed distance from a given point in the plane. The given point is called the center, $\left(h,k\right),$ and the fixed distance is called the radius, r, of the circle.
We look at a circle in the rectangular coordinate system. The radius is the distance from the center, $\left(h,k\right),$ to a point on the circle, $\left(x,y\right).$ 

To derive the equation of a circle, we can use the distance formula with the points $\left(h,k\right),$ $\left(x,y\right)$ and the distance, r. 
$\phantom{\rule{0.55em}{0ex}}d=\sqrt{{\left({x}_{2}{x}_{1}\right)}^{2}+{\left({y}_{2}{y}_{1}\right)}^{2}}$ 
Substitute the values.  $\phantom{\rule{0.55em}{0ex}}r=\sqrt{{\left(xh\right)}^{2}+{\left(yk\right)}^{2}}$ 
Square both sides.  ${r}^{2}={\left(xh\right)}^{2}+{\left(yk\right)}^{2}$ 
This is the standard form of the equation of a circle with center, $\left(h,k\right),$ and radius, r.
Standard Form of the Equation a Circle
The standard form of the equation of a circle with center, $\left(h,k\right),$ and radius, r, is
Example 11.5
Write the standard form of the equation of the circle with radius 3 and center $\left(0,0\right).$
Solution
Use the standard form of the equation of a circle  ${\left(xh\right)}^{2}+{\left(yk\right)}^{2}={r}^{2}$ 
Substitute in the values $r=3,h=0,$ and $k=0.$  ${\left(x0\right)}^{2}+{\left(y0\right)}^{2}={3}^{2}$ 
Simplify.  ${x}^{2}+{y}^{2}=9$ 
Try It 11.9
Write the standard form of the equation of the circle with a radius of 6 and center $\left(0,0\right).$
Try It 11.10
Write the standard form of the equation of the circle with a radius of 8 and center $\left(0,0\right).$
In the last example, the center was $\left(0,0\right).$ Notice what happened to the equation. Whenever the center is $\left(0,0\right),$ the standard form becomes ${x}^{2}+{y}^{2}={r}^{2}.$
Example 11.6
Write the standard form of the equation of the circle with radius 2 and center $\left(\mathrm{1},3\right).$
Solution
Use the standard form of the equation of a circle. 
$\phantom{\rule{1.3em}{0ex}}{\left(xh\right)}^{2}+{\left(yk\right)}^{2}={r}^{2}$ 
Substitute in the values.  ${\left(x\left(\mathrm{1}\right)\right)}^{2}+{\left(y3\right)}^{2}={2}^{2}$ 
Simplify.  $\phantom{\rule{0.9em}{0ex}}{\left(x+1\right)}^{2}+{\left(y3\right)}^{2}=4$ 
Try It 11.11
Write the standard form of the equation of the circle with a radius of 7 and center $\left(2,\mathrm{4}\right).$
Try It 11.12
Write the standard form of the equation of the circle with a radius of 9 and center $\left(\mathrm{3},\mathrm{5}\right).$
In the next example, the radius is not given. To calculate the radius, we use the Distance Formula with the two given points.
Example 11.7
Write the standard form of the equation of the circle with center $\left(2,4\right)$ that also contains the point $\left(\mathrm{2},1\right).$
Solution
The radius is the distance from the center to any point on the circle so we can use the distance formula to calculate it. We will use the center $\left(2,4\right)$ and point $\left(\mathrm{2},1\right)$
$\begin{array}{cccc}\text{Use the Distance Formula to find the radius.}\hfill & & & \phantom{\rule{4em}{0ex}}r=\sqrt{{\left({x}_{2}{x}_{1}\right)}^{2}+{\left({y}_{2}{y}_{1}\right)}^{2}}\hfill \\ \text{Substitute the values.}\phantom{\rule{0.2em}{0ex}}\left(\stackrel{{x}_{1},{y}_{1}}{2,4}\right),\left(\stackrel{{x}_{2},{y}_{2}}{\mathrm{2},1}\right)\hfill & & & \phantom{\rule{4em}{0ex}}r=\sqrt{{\left(\mathrm{2}2\right)}^{2}+{\left(14\right)}^{2}}\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{4em}{0ex}}r=\sqrt{{\left(\mathrm{4}\right)}^{2}+{\left(\mathrm{3}\right)}^{2}}\hfill \\ & & & \phantom{\rule{4em}{0ex}}r=\sqrt{16+9}\hfill \\ & & & \phantom{\rule{4em}{0ex}}r=\sqrt{25}\hfill \\ & & & \phantom{\rule{4em}{0ex}}r=5\hfill \end{array}$
Now that we know the radius, $r=5,$ and the center, $\left(2,4\right),$ we can use the standard form of the equation of a circle to find the equation.
$\begin{array}{cccccc}\text{Use the standard form of the equation of a circle.}\hfill & & & \phantom{\rule{2em}{0ex}}\hfill {\left(xh\right)}^{2}+{\left(yk\right)}^{2}& =\hfill & {r}^{2}\hfill \\ \text{Substitute in the values.}\hfill & & & \phantom{\rule{2em}{0ex}}{\left(x2\right)}^{2}+{\left(y4\right)}^{2}\hfill & =\hfill & {5}^{2}\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{2em}{0ex}}{\left(x2\right)}^{2}+{\left(y4\right)}^{2}\hfill & =\hfill & 25\hfill \end{array}$
Try It 11.13
Write the standard form of the equation of the circle with center $\left(2,1\right)$ that also contains the point $\left(\mathrm{2},\mathrm{2}\right).$
Try It 11.14
Write the standard form of the equation of the circle with center $\left(7,1\right)$ that also contains the point $\left(\mathrm{1},\mathrm{5}\right).$
Graph a Circle
Any equation of the form ${\left(xh\right)}^{2}+{\left(yk\right)}^{2}={r}^{2}$ is the standard form of the equation of a circle with center, $\left(h,k\right),$ and radius, r. We can then graph the circle on a rectangular coordinate system.
Note that the standard form calls for subtraction from x and y. In the next example, the equation has $x+2,$ so we need to rewrite the addition as subtraction of a negative.
Example 11.8
Find the center and radius, then graph the circle: ${\left(x+2\right)}^{2}+{\left(y1\right)}^{2}=9.$
Solution
Use the standard form of the equation of a circle. Identify the center, $\left(h,k\right)$ and radius, r. 

Center: $\left(\mathrm{2},1\right)$ radius: 3  
Graph the circle. 
Try It 11.15
ⓐ Find the center and radius, then ⓑ graph the circle: ${\left(x3\right)}^{2}+{\left(y+4\right)}^{2}=4.$
Try It 11.16
ⓐ Find the center and radius, then ⓑ graph the circle: ${\left(x3\right)}^{2}+{\left(y1\right)}^{2}=16.$
To find the center and radius, we must write the equation in standard form. In the next example, we must first get the coefficient of ${x}^{2},{y}^{2}$ to be one.
Example 11.9
Find the center and radius and then graph the circle, $4{x}^{2}+4{y}^{2}=64.$
Solution
Divide each side by 4.  
Use the standard form of the equation of a circle. Identify the center, $\left(h,k\right)$ and radius, r. 

Center: $\left(0,0\right)$ radius: 4  
Graph the circle. 
Try It 11.17
ⓐ Find the center and radius, then ⓑ graph the circle: $3{x}^{2}+3{y}^{2}=27$
Try It 11.18
ⓐ Find the center and radius, then ⓑ graph the circle: $5{x}^{2}+5{y}^{2}=125$
If we expand the equation from Example 11.8, ${\left(x+2\right)}^{2}+{\left(y1\right)}^{2}=9,$ the equation of the circle looks very different.
$\begin{array}{cccccc}& & & \hfill {\left(x+2\right)}^{2}+{\left(y1\right)}^{2}& =\hfill & 9\hfill \\ \text{Square the binomials.}\hfill & & & \hfill {x}^{2}+4x+4+{y}^{2}2y+1& =\hfill & 9\hfill \\ \begin{array}{c}\text{Arrange the terms in descending degree order,}\hfill \\ \text{and get zero on the right}\hfill \end{array}\hfill & & & \hfill {x}^{2}+{y}^{2}+4x2y4& =\hfill & 0\hfill \end{array}$
This form of the equation is called the general form of the equation of the circle.
General Form of the Equation of a Circle
The general form of the equation of a circle is
If we are given an equation in general form, we can change it to standard form by completing the squares in both x and y. Then we can graph the circle using its center and radius.
Example 11.10
ⓐ Find the center and radius, then ⓑ graph the circle: ${x}^{2}+{y}^{2}4x6y+4=0.$
Solution
We need to rewrite this general form into standard form in order to find the center and radius.
Group the xterms and yterms. Collect the constants on the right side. 

Complete the squares.  
Rewrite as binomial squares.  
Identify the center and radius.  Center: $\left(2,3\right)$ radius: 3 
Graph the circle. 
Try It 11.19
ⓐ Find the center and radius, then ⓑ graph the circle: ${x}^{2}+{y}^{2}6x8y+9=0.$
Try It 11.20
ⓐ Find the center and radius, then ⓑ graph the circle: ${x}^{2}+{y}^{2}+6x2y+1=0.$
In the next example, there is a yterm and a ${y}^{2}$term. But notice that there is no xterm, only an ${x}^{2}$term. We have seen this before and know that it means h is 0. We will need to complete the square for the y terms, but not for the x terms.
Example 11.11
ⓐ Find the center and radius, then ⓑ graph the circle: ${x}^{2}+{y}^{2}+8y=0.$
Solution
We need to rewrite this general form into standard form in order to find the center and radius.
Group the xterms and yterms.  
There are no constants to collect on the right side. 

Complete the square for ${y}^{2}+8y.$  
Rewrite as binomial squares.  
Identify the center and radius.  Center: $\left(0,\mathrm{4}\right)$ radius: 4 
Graph the circle. 
Try It 11.21
ⓐ Find the center and radius, then ⓑ graph the circle: ${x}^{2}+{y}^{2}2x3=0.$
Try It 11.22
ⓐ Find the center and radius, then ⓑ graph the circle: ${x}^{2}+{y}^{2}12y+11=0.$
Media
Access these online resources for additional instructions and practice with using the distance and midpoint formulas, and graphing circles.
Section 11.1 Exercises
Practice Makes Perfect
Use the Distance Formula
In the following exercises, find the distance between the points. Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.
$\left(\mathrm{4},\mathrm{3}\right)$ and $\left(2,5\right)$
$\left(\mathrm{7},\mathrm{3}\right)$ and $\left(8,5\right)$
$\left(\mathrm{1},3\right)$ and $\left(5,\mathrm{5}\right)$
$\left(\mathrm{8},\mathrm{2}\right)$ and $\left(7,6\right)$
$\left(\mathrm{1},\mathrm{2}\right)$ and $\left(\mathrm{3},4\right)$
$\left(\mathrm{4},\mathrm{5}\right)$ and $\left(7,4\right)$
Use the Midpoint Formula
In the following exercises, ⓐ find the midpoint of the line segments whose endpoints are given and ⓑ plot the endpoints and the midpoint on a rectangular coordinate system.
$\left(\mathrm{2},\mathrm{6}\right)$ and $\left(6,\mathrm{2}\right)$
$\left(\mathrm{3},\mathrm{3}\right)$ and $\left(6,\mathrm{1}\right)$
Write the Equation of a Circle in Standard Form
In the following exercises, write the standard form of the equation of the circle with the given radius and center $\left(0,0\right).$
Radius: 9
Radius: $\sqrt{5}$
In the following exercises, write the standard form of the equation of the circle with the given radius and center
Radius: 10, center: $\left(\mathrm{2},6\right)$
Radius: $1.5,$ center: $\left(\mathrm{5.5},\mathrm{6.5}\right)$
For the following exercises, write the standard form of the equation of the circle with the given center with point on the circle.
Center $\left(6,\mathrm{6}\right)$ with point $\left(2,\mathrm{3}\right)$
Center $\left(\mathrm{5},6\right)$ with point $\left(\mathrm{2},3\right)$
Graph a Circle
In the following exercises, ⓐ find the center and radius, then ⓑ graph each circle.
${\left(x2\right)}^{2}+{\left(y3\right)}^{2}=9$
${\left(x+2\right)}^{2}+{\left(y5\right)}^{2}=4$
${\left(x1\right)}^{2}+{y}^{2}=36$
${\left(x1\right)}^{2}+{\left(y3\right)}^{2}=\frac{9}{4}$
${x}^{2}+{y}^{2}=49$
$6{x}^{2}+6{y}^{2}=216$
In the following exercises, ⓐ identify the center and radius and ⓑ graph.
${x}^{2}+{y}^{2}6x8y=0$
${x}^{2}+{y}^{2}+12x14y+21=0$
${x}^{2}+{y}^{2}10y=0$
${x}^{2}+{y}^{2}14x+13=0$
Writing Exercises
Is a circle a function? Explain why or why not.
In your own words, explain the steps you would take to change the general form of the equation of a circle to the standard form.
Self Check
ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
ⓑ If most of your checks were:
…confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.
…with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Who can you ask for help? Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?
…no  I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.