Lesson Graphing Circles in Color

Algebra ->  Algebra  -> Graphs -> Lesson Graphing Circles in Color      Log On

Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!
Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!

   

This Lesson (Graphing Circles in Color) was created by by rapaljer(4551) About Me : View Source, Show
About rapaljer: Retired Professor of Mathematics from Seminole State College of Florida after 36 years.

This text was imported from http://www2.seminolestate.edu/rrapalje/Math%20in%20Living%20Color/College%20Living%20Color/3204%20Circle%20College%20Alg.htm by its author.

2.04  The Circle

College Algebra: One Step at a Time,  Page 234 - 241:   #18, 19, 26, 28, 30, 39, 43

Dr. Robert J. Rapalje

Seminole State College of Florida

Altamonte Springs Campus

 

To see Section 2.04, with detailed explanations, examples, exercises, and answers, click here!

 

Circle Summary

represents the equation of a circle with center at

 and of radius r.

also represents a circle (or a point or no solution!)

  

 

p. 237  # 18.      

Solution:  Before you ever start the problem, you know that this looks like a CIRCLE!  The best way to find the center and radius is by completing the square.   Since you already  have coefficients of  and , you should begin by re-writing the equation with the x terms together and y terms together, leaving a space to complete the square:

When you complete the square, remember that you take half of the coefficient, and square.  Half of  is and is .  Half of  is , and is .

 

   

                         

This is a circle with center at and with radius .

To graph this circle, start at the origin and count 4 units to the left, then up 3 units.  This is the center of the circle.  Now, the radius is  so measure 8 units in each direction from the center.  The graph should look like this.  

                                   

Circle with center at and with radius .

 

p. 238.  # 19      

Solution:  As in #18, this is a completing the square problem.   Begin by re-writing the equation with the x terms together and y terms together, leaving a space to complete the square.  Also, you should add +48 to each side of the equation:

When you complete the square, remember that you take half of the coefficient, and square.  Half of  is and is .  Half of  is , and is .

     

    

                         

This is a circle with center at and with radius .

To graph this circle, start at the origin and count 6 units to the right, then down 4 units.  This is the center of the circle.  Now, the radius is  so measure 10 units in each direction from the center.  The graph should look like this.  

                               

Circle with center at and with radius .

 

 

p. 238.  # 26.      

Solution:  Before you ever start the problem, you know that this looks like a CIRCLE!  The most effective way to find the center and radius is by completing the square.   However, before you can complete the square, you need to have coefficients of  and .  The best way to accomplish this is to divide both sides of the equation by .

                                   

                                      

Rearrange the terms with the x terms together, and the y terms together, and leave spaces to complete the square for each variable.

                                      

          

When you complete the square, remember that you take half of the coefficient, and square.  Half of  is and is .  Half of  is , and is .

         

                     

                                    

                                    

                                    

This is a circle with center at and with radius .

 

 

 

Page 238. # 28. 

Solution:    

Before you ever start the problem, you know that this looks like a CIRCLE!  The most effective way to find the center and radius is by completing the square.   However, before you can complete the square, you need to have coefficients of  and .  The best way to accomplish this is to divide both sides of the equation by .

Rearrange the terms with the x terms together, and the y terms together, and leave spaces to complete the square for each variable.

When you complete the square, remember that you take half of the coefficient, and square.  Half of  is and is

Half of  is , and is .

          

                      

This is a circle with center at and with radius .

To graph this circle, start at the origin and count 2.5 units to the right, then up 1.5 units.  This is the center of the circle.  Now, the radius is or 1.5 so measure 1.5 units in each direction from the center.  The graph should look like this. 

                     

Circle with center at and of radius .

p. 239.  # 30.      

Solution:  Before you ever start the problem, you know that this looks like a CIRCLE!  The most effective way to find the center and radius is by completing the square.   However, before you can complete the square, you need to have coefficients of  and .  The best way to accomplish this is to divide both sides of the equation by .

Rearrange the terms with the x terms together, and the y terms together, and leave spaces to complete the square for each variable.

When you complete the square, remember that you take half of the coefficient, and square.  Half of  is and is .  Half of  is , and is .

This is a circle with center at and with radius , so

To graph this circle, start at the origin and count 2.5 units to the right, then down 0.5 units.  This is the center of the circle.  Now, the radius is or 0.7 so measure about .7 units(less than 1 unit) in each direction from the center.  The graph should look like this. 

 

Circle with center at and with radius .

 

 

p. 240.  # 39.    Find the equation of a circle with center at  and passing through .

Solution:  In order to find the equation of a circle, you must know the  center and the radius of the circle.     In this case, you are given the center, but not the radius of the circle.   However, if you know the center and a point that is ON the circle, then the distance between these two points is the radius.  You must find the distance between the two given points by the distance formula

 

Distance Formula

The distance between two points  and is given by the formula

 

Now, the distance between the two points given in this problem, and therefore the radius of the circle, is

 Radius =

Radius =

Radius =

Radius =

Therefore the equation of the circle with center at  and of radius  is given by

As a check (no extra charge!!), you might want to sketch the graph, and see if it does indeed pass through the point .  As you can see, it does!!

 

                     

Final Answer: 

 

 

 

Return to main page        Math in Living C O L O R !!

     Return to Intermediate Algebra page  

 Return to College Algebra page

 

Dr. Robert J. Rapalje Altamonte Springs Campus
Contact me at:   rapaljer@seminolestate.edu
Phone number:  NONE Retired!!
OFFICE:          NONE  
Copyright © Seminole State College of Florida, 1997



This lesson has been accessed 7407 times.