Conic Section And Analytical Geometery, Study Guides, Projects, Research of Mathematics

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12.1 Parabolas

12.2 Circles

12.3 Ellipses

12.4 Hyperbolas

Conic

Sections

Chapter

S

mooth, regular curves are common in the world

around us. The orbit of the space shuttle around the

Earth is one kind of smooth curve; another is the

shape of domed buildings such as the United States Capitol.

Television receiving dishes are a very common curved

shape, as are the curved rearview mirrors on your car.

Mathematically, however, an interesting property of all the

curves just mentioned is that they are expressed as qua-

dratic equations in two variables.

We discussed quadratic equations in Chapter 5. In the

current chapter, we extend that discussion to other forms of

equations and their related graphs, called conic sections:

parabolas, circles, ellipses, and hyperbolas. We will describe

each curve and analyze the equation used to graph it. We

will also discuss properties that make these curves so use-

ful in so many different areas, from engineering to archi-

tecture to astronomy. We conclude the chapter with a

project that further explores the various conic sections we

have studied.

12.1 APPLICATION

A satellite dish receiver is in the shape of a parabola. A cross section of the dish shows a diameter of 13 feet at a distance of 2. feet from the vertex of the parabola. Write an equation for the parabola. After completing this section, we will discuss this application fur- ther. See page 000.

12.1.

12.1 • Parabolas 949

To write a quadratic equation in the form

• Isolate the x -terms to one side of the equation.
• Factor out the leading coefficient.
• Add the value needed to complete the square to both sides of the equation.
• Rewrite the trinomial as a binomial squared.
• Solve the equation for y.

y  a 1 x  h 22  k,

EXAMPLE 1 Write the quadratic equations in the form Identify and k. a. b.

Solution a. Isolate the x -terms. Add to both sides. Write the trinomial as a binomial squared. Solve for y****. In the equation and b. Isolate the x -terms. Factor out the leading coefficient,

Add or to both sides. Write the trinomial as a binomial squared. Solve for y****. In the equation or a  2, h  4,and k  3. •

y   21 x  422  3, y   23 x  1  4 2 4^2  1  32 ,

y   21 x  422  3

y  3   21 x  422

y  35  3  2116 2 4   21 x^2  8 x  162  2 1^82 2^2 ,  2 1 16 2 ,

 2.

y  35   21 x^2  8 x 2

y  35   2 x^2  16 x

y   2 x^2  16 x  35

y  1 x  222  3, a  1, h  2, k  3.

y  1 x  222  3

y  3  1 x  222

y  7  4  x^2  4 x  4 1  24 2^2  4

y  7  x^2  4 x

y  x^2  4 x  7

y  x^2  4 x  7 y   2 x^2  16 x  35

y  a 1 x  h 22  k. a, h,

HELPING HAND Note that the leading coefficient a is also the same value as the real number a in our new equation.

12.1.1 Checkup

In exercises 1 and 2, write the quadratic equations in the form Identify a , h , and k.

1. y  x^2  6 x  7 2. y  3 x^2  12 x  5

y  a 1 x  h 22  k.

3. When you rewrite a quadratic equation in the form is the new equation equivalent to the original equation? Explain.

y  a 1 x  h 22  k,

12.1.2 Understanding the Effects of the Constants In Chapter 5, we described the graph of a quadratic equation in two variables, as a parabola. We discovered that the values of the real numbers a, b,and c affected the shape of the graph.

y  f 1 x 2  ax^2 ^ bx^ ^ c,

(

950 Chapter 12 • Conic Sections

Discovery 1

Effect of the Real Numbers h and k of a Quadratic Equation on a Parabola

1. Sketch the graphs of the following quadratic equations of the form where and Label the vertex of each graph. 2. Write a rule for determining the y -coordinate of the vertex of a parabola from the equation of the parabola. 3. Sketch the graphs of the following quadratic equations of the form where and Label the vertex of each graph. 4. Write a rule for determining the x -coordinate of the vertex of a parabola from the equation of the parabola. 5. Sketch the graph of the following quadratic equations of the form where Label the vertex of each graph. 6. Write a rule for determining the coordinates of the vertex of a parabola from the equation of the parabola.

y  1 x  322  2 y  1 x  322  2

y  a 1 x  h 22  k, a  1.

y  1 x  322 y  1 x  322

y  a 1 x  h 22  k, a  1 k  0.

y  x^2  2 y  x^2  2

y  a 1 x  h 22  k, a  1 h  0.

The vertex of the parabola is Since the x -coordinate of the vertex is h , the axis of symmetry is the graph of the vertical line The graph of a qua- dratic equation of the form is called a vertical parabola , because its axis of symmetry is a vertical line. Remember that the real number a is also the leading coefficient of the equa- tion Therefore, it also affects the shape of the graph.

SUMMARY OF THE EFFECTS OF THE REAL NUMBERS a , h , AND k OF A QUADRATIC EQUATION ON A VERTICAL PARABOLA The real numbers and k of a quadratic equation in the form affect the graph of the equation. If then the graph is concave upward (opens upward). If then the graph is concave downward (opens downward). If then the graph is narrower than it would be if If then the graph is wider than it would be if The vertex of the graph is The axis of symmetry is the line graphed by x  h.

1 h, k 2.

0 a 0 6 1, a  1.

0 a 0 7 1, a  1.

a 6 0,

a 7 0,

y  a 1 x  h 22  k

a, h,

y  ax^2  bx  c.

y  a 1 x  h 22  k

x  h.

1 h, k 2.

EXAMPLE 2 Determine the vertex and axis of symmetry for the graph of each equation. Describe the graph, but do not draw it. a. y  21 x  422  3 b. y  x^2  4 x  8

If the quadratic equation is written in the form do the values of the real numbers and k affect the shape of the parabola? To find out, complete the following set of exercises.

a, h,

y  a 1 x  h 22  k,

952 Chapter 12 • Conic Sections

Therefore, and The vertex is or The axis of symmetry is the line The graph opens upward, because The graph is narrower than it would be if because The y -intercept is the point on the graph where Substitute 0 for x and solve for y.

The y -intercept is The x -intercept is the point on the graph where Therefore, substi- tute 0 for y and solve for x.

The x -intercepts are about and

Calculator Check

x 

x  4 ;

x  4 ; A

x  4 ; A

x  4  ; A

1 x  422 

21 x  422  3

0  21 x  422  3

y  21 x  422  3

y  0.

y  29

y  32  3

y  21162  3

y  210  422  3

y  21 x  422  3

x  0.

a  1, 0 a 0  02 0 7 1.

a  2 7 0.

x  4.

1 h, k 2 , 1 4,  32.

a  2, h  4, k  3.

x

y

 4

4

8

12

32 28 24

16

20

 1 1 2 3 5 6 7 8 9

(0, 29)

(2.78, 0)

(5.22, 0)

(4, 3)

y  2( x  4)^2  3

x  4

Y1  2( x  4)^2  3

1 10, 10, 10, 10 2

12.1.

x

y

 10

 8

 6

 4

 2

10 8 6 4 2  10  8  6  4  2 2 4 810

y   x  4

y   x  4

12.1 • Parabolas 953

12.1.3 Checkup

1. Graph the vertical parabola for 2. Name some points that are useful to locate in graphing a vertical parabola, and explain why they are useful.

y  ^23 1 x  522  2.

12.1.4 Graphing a Horizontal Parabola Some parabolas open left or right. In such a case, the parabola has a hori- zontal axis of symmetry and is called a horizontal parabola. For example, graph the horizontal parabola In order to graph this parabola, we will solve for y.

We will graph two curves, and on the same coordinate plane. To graph a horizontal parabola on your calculator, graph two curves. Note that a horizontal parabola does not represent a function, because it does not pass the vertical-line test. The form of a quadratic equation that will graph a horizontal parabola is To write an equation in this form, we complete the square as we did for a vertical parabola, only this time for the variable y instead of x. In addition, just as the real numbers a , h , and k affect a vertical parabola, they also affect a horizontal parabola.

SUMMARY OF THE EFFECTS OF THE REAL NUMBERS

a , h , AND k OF A QUADRATIC EQUATION ON A HORIZONTAL PARABOLA The real numbers a , h , and k of a quadratic equation in the form affect its graph. If then the graph opens to the right. If then the graph opens to the left. If then the graph is narrower than it would be if If then the graph is wider than it would be if The vertex of the graph is The axis of symmetry is the line graphed by y  k.

1 h, k 2.

0 a 0 6 1, a  1.

0 a 0 7 1, a  1.

a 6 0,

a 7 0,

x  a 1 y  k 22  h

x  a 1 y  k 22  h.

y  1 x  4 y   1 x  4,

y  ; 1 x  4

y^2  x  4

x  4  y^2

x  y^2  4

x  y^2  4.

Y1  x  4

Y2   x  4

1 10, 10, 10, 10 2

EXAMPLE 4 Graph the horizontal parabolas. a. b.

Solution a. The equation is written in the form Therefore, and The vertex is or The axis of symmetry is the graph of or The graph opens to the right, because a  1 7 0.

y  k, y  3.

1 h, k 2 , 1 5, 3 2.

a  1, h  5, k  3.

x  a 1 y  k 22  h.

x  1 y  322  5

x  1 y  322  5 x   2 y^2  4 y  5

12.1.

12.1 • Parabolas 955

Calculator Check Solve the equation for y. First, we complete the square as in the previous so- lution, obtaining

Isolate the y terms to one side of the equation.

Divide both sides by

Principle of square roots

Solve for y****.

y   1 ; (^) A •

x  3 2

A

x  3 2

 y

A

x  3 2

 y  1

 2.

x  3 2

 1 y  122

x  3   21 y  122

x   21 y  122  3

x   21 y  122  3.

12.1.4 Checkup

In exercises 1 and 2, graph the horizontal parabola for each relation.

1. 2. x  

1 2 x  21 y  122  3 y^2  4 y  7

3. How does a horizontal parabola differ from a vertical one? 4. How can the constants in the equation for a horizontal parabola help you graph the equation?

x

y

 10

 8

 6

 4

 2

10 8 6 4 2  10  8  6  2 2 4 6 8 10

x  2( y  1)^2  3

(3, 1)^ y^ ^ ^1

(5, 0)

1 10, 10, 10, 10 2

Y1   1  _______ x^ 2 ^3

Y2   1  _______ x^ 2 ^3

12.1.5 Writing Quadratic Equations, Given the Vertex and a Point on the Graph Earlier, we learned that although two points determine a straight line, you need three points to determine a curve. Thus, an infinite number of parabo- las can be drawn through any two given points. However, if we know that one of these points is the vertex and if we know that the parabola is vertical or horizontal, then we can write an equation for the specific parabola that passes through these points.

956 Chapter 12 • Conic Sections

EXAMPLE 5 a. Write an equation of a vertical parabola with a vertex of and passing through the point b. Write an equation of a horizontal parabola with a vertex of and a y -intercept of

Solution a. Since the vertex is it follows that and Also, we know that the point is a solution of the equation. We will substitute for x and 4 for y , as well as 2 for h and 6 for k , in the equation and solve for a.

Substitute.

Now, we write an equation using the known values for a , h , and k.

The graph of the equation is a vertical parabola with a vertex of and passing through the point b. First, we substitute values for h , k , x , and y. Given the vertex we know that and We use the coordinates of the y -intercept for x and y , and Then we solve for a.

Substitute.

Write an equation using and

The graph of the equation is a horizontal parabola with a vertex of 1 1, 1 2 and a y -intercept of 1 0, 2 2. •

x  1 y  122  1

x  1 y  122  1

x  11 y  122  1  12

x  a 1 y  k 22  h

a  1, h  1, k  1.

a  1

0  a  1

0  a 1122  1

0  a 12  122  1  12

x  a 1 y  k 22  h

x  0 y  2.

h   1 k  1.

y  ^29 1 x  222  6

y  

1 x  222  6

y  a 1 x  h 22  k

a  

4  9 a  6

4  a 1  322  6

4  a3 1 12  242  6

y  a 1 x  h 22  k

y  a 1 x  h 22  k

1 2, 6 2 , h  2 k  6.

12.1.5 Checkup

1. Write an equation of a vertical parabola with a vertex of and passing through 2. Write an equation of a horizontal parabola with a vertex of 1 1,  22 and passing through 1 9,  42.

1 6, 2 2 1 3, 5 2.

3. When you are asked to write an equation of a parabola and are given points on its graph to do so, is it important to know whether the parabola is a vertical parabola or a horizontal parabola? Explain.

25 ft

15 ft

958 Chapter 12 • Conic Sections

12.1.6 Checkup

1. An arched underpass has the shape of a parabola. A road passing under the arch is 25 feet wide, and the maximum height of the arch is 15 feet. Write an equation for the par- abolic arch.

APPLICATION A satellite dish receiver is in the shape of a parabola. A cross section of the dish shows a diameter of 13 feet at a distance of 2. feet from the vertex of the parabola. Write an equation for the parabola.

Discussion According to the figure, the horizontal parabola has a vertex at the origin. If the parabola is 13 feet in diameter at a distance along the axis of 2.5 feet from the vertex, the radius is 6.5 feet. We can thus label two points on the parabola as and

Substitute 0 for h , 0 for k , 2.5 for x , and 6.5 for y in the equation for a horizontal parabola, and solve for a.

Since our object is not modeled by the complete graph, we need to limit the x -values by restricting the domain. An equation for the parabola is where 6.5  y  6.5.

x  16910 y^2 ,

a 

2.5  42.25a

2.5  a 1 6.5  022  0

x  a 1 y  k 22  h

(0, 0) x

y

 10

 8

 6

 4

 2

10 8 6 4 2  10  8  6  4  2 2 4 6 810

(2.5, 6.5)

(2.5, 6.5)

13 ft

x

y

2.5 ft

2. A soup bowl has a cross section with a parabolic shape, as shown in the figure. The bowl has a diameter of 8 inches and is 2.5 inches deep. Write an equation for its shape.

8 in.

2.5 in.

12.1 • Parabolas 959

12.1 Exercises

Write each equation in the form

**1. 2. 3.

6. 11. 12.**

Determine the vertex and axis of symmetry, and describe the graph, for each equation. Do not graph the equation.

**13. 14. 15.

20. 29. 30.**

Graph each parabola.

**31. 32. 33.

41. 56. 57.**

58.

Write an equation of the vertical parabola with the given vertex and passing through the given point.

59. Vertex at and passing through 60. Vertex at and passing through 61. Vertex at and passing through 62. Vertex at and passing through 63. Vertex at and passing through 64. Vertex at and passing through 65. Vertex at and passing through 66. Vertex at and passing through

In exercises 67–72, write an equation of the horizontal parabola with the given vertex and passing through the given point.

67. Vertex at and passing through 68. Vertex at and passing through 69. Vertex at and passing through 70. Vertex at and passing through 71. Vertex at 1 8, 2 2 and passing through 1 2,  32. 72. Vertex at 1 0,  12 and passing through 1 6, 1 2.

1 5,  42 1 7,  22. 1 4,  32 1 10,  62.

1 4, 3 2 1 6, 2 2. 1 3,  52 1 5,  62.

1 0,  22 1 2, 0 2. 1 4,  52 1 6, 1 2.

1 4,  52 1 0, 3 2. 1 3, 11 2 1 3,  132.

1 1, 4 2 1 3, 12 2. 1 2,  12 1 1, 26 2.

1 3,  22 1 5, 2 2. 1 5, 3 2 1 3,  12.

y  1.5x^2  3 x  4.

x  x  0.4y^2  0.8y  2. 3 4 y  y^2  3 y  2 1 3 x^2  4 x  7

x  3 y^2  6 y  5 x   2 y^2  16 y  22 y   4 x^2  16 x  7

x  0.8 1 y  122  2 x  1.5 1 y  222  2.5 y  2 x^2  12 x  23

x  1 4

x   1 y  222 1 2

x  2 y^2  3 1 y  422  7

x  1 y  322  1 x   31 y  122  8 x  21 y  522

y  0.5 1 x  222 y  0.2 1 x  122  2 y  1.4x^2  3

y  y  1.6x^2

4 5 y   x^2  3

2 3 1 x  322  4

y   3 4 y  1 x  222  3 1 2 y   41 x  122  1 1 x  222  5

y  31 x  522  1 y  1 x  222  3 y   1 x  222  2

y  y  1.2x^2  2.4x  4.6 y  0.8x^2  8 x  9. 2 3 x^2  8 x  35

y  

1 2 y  4 x^2  24 x  25 y   3 x^2  12 x x^2  6 x  7

y  3.7 1 x  122  4.8 y  4.2 1 x  1222  5 y  0.75 1 x  522  6

y  y  0.8 1 x  922  1

5 7 y   1 x  1422  7

1 4 1 x  322  10

y   1 5 y  1 x  1022  11 2 5 y  41 x  522  7 1 x  522  7

y  31 x  1222  5 y   51 x  422  9 y   21 x  1322  5

y  y  1.5x^2  6 x  2.4 y  2.5x^2  20 x  34. 2 3

x^2  8 x  7

y 

1 2 y  x^2  20 x  103 y  2 x^2  44 x  227 x^2  4 x  13

y   2 x^2  28 x  89 y   3 x^2  42 x  157 y  5 x^2  30 x  32

y  2 x^2  36 x  165 y   4 x^2  24 x  34 y  3 x^2  48 x  197

y  a 1 x  h 22  k.

73. A steel bridge has an arch shaped like a parabola. The maximum height of the arch is 35 feet, and the road on the bridge spans a distance of 90 feet. Write an equation for the parabolic arch. 74. A wooden bridge has an arch shaped like a parabola, with a maximum height of 20 feet. The road on the bridge stretches a distance of 75 feet. Write an equation for the parabolic arch of the bridge.

12.2.

center radius

12.2 • Circles 961

12.2 Circles

OBJECTIVES 1 Graph circles with the center at the origin.

2 Graph circles with the center not at the origin. 3 Write equations of circles. 4 Model real-world situations by using circles.

A circle is the set of points in a plane that are equidistant from a given point, called the center. The radius (plural, radii ) of the circle is the distance be- tween each of its points and the center.

APPLICATION Many railroad viaducts are constructed in the shape of a semicircle. A stone-arch rail- road viaduct at Rockville, Pennsylvania, over the Susquehanna River is made of 48 semicircular arches, each with a span of 70 feet. Use Figure 12.1 to write equations that model each of the first two arches. After completing this section, we will discuss this application further. See page 000. Figure 12.

10 ft 70 ft 10 ft 70 ft

y

x

12.2.1 Graphing Circles with the Center at the Origin Since a circle is defined as a set of points equidistant from the center, we can use the distance formula to determine an equation of a circle. We begin with a circle having its center at the origin passing through a point and having a radius r. Distance formula Substitute. Square both sides.

CENTER–RADIUS FORM OF THE EQUATION OF A CIRCLE

WITH CENTER AT THE ORIGIN

The center–radius form of the equation of a circle with its center at the origin and radius r is where

We can now graph a circle, using this equation.

To graph a circle with the center at the origin,

• Locate and label the origin as the center of the circle.
• Locate and label the two x -intercepts and
• Locate and label the two y -intercepts and
• Sketch the circle containing the preceding two points.

1 0, r 2 1 0, r 2.

1 r, 0 2 1 r, 0 2.

x^2  y^2  r^2 , r 7 0

or x^2  y^2  r^2

r^2  x^2  y^2

r  21 x  022  1 y  022

D  21 x 2  x 122  1 y 2  y 122

C 1 0, 0 2 , P 1 x, y 2 ,

x

y

P ( x , y )

C (0, 0)

r

962 Chapter 12 • Conic Sections

x

y

 5

 4

 2

 1

5 4

2 1  5  4  2  1 1 2 4 5

C (0, 0)

x^2  y^2  10

(10, 0)

(0, 10) r   10

(0, 10)

(10, 0)

Y1  10  x^2

Y2   10  x^2 1 9.4, 9.4, 2, 6.2, 6.2, 2 2

EXAMPLE 1 Graph. a. b.

Solution Calculator Check a. Rewrite the constant as a square.

Since it follows that The circle has its center at the origin and a radius of 3.

r^2  9  32 , r  3.

x^2  y^2  32

x^2  y^2  9

x^2  y^2  9 x^2  y^2  10

b. Calculator Check Rewrite the constant as a square.

Since it follows that r  110  3.16.

r^2  10  1 11022 ,

x^2  y^2  1 11022

x^2  y^2  10

x

y

 5

 4

 2

 1

5 4

2 1  5  4  2  1 1 2 4 5

(0, 3)

(0, 3)

(3, 0) (3, 0)

C (0, 0)

x^2  y^2  9 r  3

To graph the equation on your calculator, first solve for y.

Subtract from both sides.

Principle of square roots The graph does not represent a function and must be entered as two functions. The graph of is a semicircle above the x -axis, and the graph of is a semicircle below the x -axis.

y   29  x^2

y  29  x^2

y  ; 29  x^2

x^2

y^2  9  x^2

x^2  y^2  9

1 4.7, 4.7, 3.1, 3.1 2

Y1  29  x^2

Y2   29  x^2

To graph the equation on your calculator, first solve for y.

Subtract from both sides.

Principle of square roots

y  ; 210  x^2

x^2

y^2  10  x^2

x^2  y^2  10

To graph this equation on your calculator, you must solve for y.

Isolate the y -terms.

Solve for y****.

y  ; 211  1 x  322

y^2  11  1 x  322

1 x  322  y^2  11

Solution Calculator Check a.

We have the equation for a circle with and The center of the circle, 1 h, k 2 ,is 1 3, 1 2 ,and the radius is 2.

h  3, k  1, r  2.

1 x  322  1 y  122  2 2

1 x  322  1 y  122  4

x

y

 5

 4

 3

 2

 1

5 4 3 2 1  5  4  3  2  1 1 2 3 4 C (3, 0)

( x  3)^2  y^2  11

r   11

Y1  11  ( x  3)^2

Y2   11  ( x  3)^2 1 9.4, 9.4, 2, 6.2, 6.2, 2 2

x

y

 5

 4

 3

 2

 1

5 4 3

1  5  4  3  2  1 1 5

C (3, 1)

(3, 3) (1, 1)

(3, 1)

(5, 1)

( x  3)^2  ( y  1)^2  4

r  2

964 Chapter 12 • Conic Sections

Y1  1  4  ( x  3)^2

Y2  1  4  ( x  3)^2 1 9.4, 9.4, 2, 6.2, 6.2, 2 2

If we can write an equation in the center–radius form, we can sketch the cir- cular graph. In Example 3, we must first write the equation in center–radius form by completing the square twice before we graph.

EXAMPLE 3 Sketch the graph of the equation

Solution

Group the x -terms and the y -terms. 1 x^2  2 x 2  1 y^2  4 y 2  3

x^2  y^2  2 x  4 y  3

x^2  y^2  2 x  4 y  3.

b.

Center–radius form for a circle Therefore, and The center is and the radius is 111  3.32.

h  3, k  0, r  1 11.

3 x  1  3 2 4^2  1 y  022  1 11122

1 x  322  y^2  11

To graph this equation on your calculator, you must solve for y.

Isolate the y -terms.

Principle of square roots

Solve for y****.

y  1 ; 24  1 x  322

y  1  ; 24  1 x  322

1 y  122  4  1 x  322

1 x  322  1 y  122  4

12.2.

Complete both squares. For the x -terms, add 1 to both sides because For the y -terms, add 4 to both sides because

Write each trinomial as a binomial square, and simplify the right side of the equation.

Write the equation in center–radius form. Remember, The center of the circle is and the radius is

Calculator Check Solve the equation for y. Complete the squares as in the previous solution.

Isolate the y -terms. Principle of square roots y  2 ; 28  1 x  (^12) • 2

y  2  ; 28  1 x  122

1 y  222  8  1 x  122

1 x  122  1 y  222  8

x^2  y^2  2 x  4 y  3

3 x  1  1 2 4^2  1 y  222  121222 1 8  2 1 2.

1 x  122  1 y  222  8

1 x^2  2 x  12  1 y^2  4 y  42  3  1  4

Y1  2  8  ( x  1)^2

Y2  2  8  ( x  1)^2 1 9.4, 9.4, 2, 6.2, 6.2, 2 2

x

y

 5

 4

 3

 2

5 4 3 2 1  5  4  3 1 2 3 4 5

C (1, 2) (^) x (^2)  y (^2)  2 x  4 y  3

r  2  2

12.2 • Circles 965

12.2.2 Checkup

In exercises 1–3, sketch the graph of each equation. Check, using your calculator.

1. 2. 3. 4. Explain how the constants in the center–radius form of the equation of a circle help you to graph the equation.

1 x  222  1 y  222  4 1 x  222  1 y  222  9 x^2  y^2  6 x  4 y  3

12.2.3 Writing Equations of Circles If we can determine the center and radius of a circle, we can use the center– radius form to write an equation of the circle.

EXAMPLE 4 Write an equation of each circle, given the following information: a. Center at the origin and radius 7 b. Center at and radius 4 c. Center at and passing through the point

Solution a. Since the circle has its center at the origin, use the equation and substitute 7 for r.

The graph of the equation is a circle with the center at the ori- gin and radius 7.

x^2  y^2  49

x^2  y^2  49

x^2  y^2  72

x^2  y^2  r^2

x^2  y^2  r^2

x

y

 10

 8

 6

 4

 2

10 8 6 4 2  10  8  6  4  2 2 4 6 810 C (0, 0)

x^2  y^2  49

r  7