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posted by  Cowboy15 on 11/17/2007 8:08:56 PM  |  status: Closed  

Algebra

Course Textbook Chapter Problem
Algebra N/A N/A N/A
Question Details:
Complete the square and write the equation in standard form. Then give the center and radius of the circle.
 
 
 
 
 
Can you show each step. Please
 
 
 
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Oracle
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posted by RPGillespie on 11/17/2007 8:20:59 PM  |  status: Live
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Cowboy15's comment:
"great work"
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Don't forget to rate!

The answers are in red:

..........50 = 49 + 1



.........................(standard form)

To find the center of the circle, set the insides of each square to 0, and solve:






Center of the circle is located at:


So, the standard form of a circle is:


In this case,

......................(radius)
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Tags: Math, Algebra
Oracle
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posted by zanthymx on 11/17/2007 10:00:09 PM  |  status: Live
Asker's Rating: Helpful   
Cowboy15's comment:
"thank-you"
Response Details:

x2 + y2 - 14*x - 2*y + 50  =  4     (<--- original problem equation)

x2 - 14*x + y2 - 2*y + 50  =  4      (<--- rearrange "x" & "y" terms)

x2 - 14*x + y2 - 2*y + 50 - 49  =  4 - 49    (<--- complete the "y" square)

x2 - 14*x + y2 - 2*y + 1  =  -45

x2 - 14*x + 49 + y2 - 2*y + 1  =  -45 + 49    (<--- complete the "x" square)

x2 - 14*x + 49 + y2 - 2*y + 1  =  4

{x2 - 14*x + 49} + {y2 - 2*y + 1}  =  4     (<--- collect terms)

(x - 7)2 + (y - 1)2  =  (2)2   (<--- standard form for circle)

---->    {Center @ x=7 & y=1}
---->    {CENTER AT}  =  (7, 1)
---->    {RADIUS}  =  2

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Tags: Math, Algebra
Mentor
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posted by misscrt on 11/18/2007 6:40:49 AM  |  status: Live
Asker's Rating: Helpful   
Cowboy15's comment:
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Response Details:
Given:  x^2 + y^2 -14x -2y +50 = 4

First, rearrange and group the terms as shown:

            (x^2 - 14x) + (y^2 - 3y) = 4 - 50
            (x^2 - 14x) + (y^2 - 2y) = - 46

Next, complete the square of (x^2 -14x) and (y^2 - 2y):

 (x^2 - 14x + 49) + (y^2 - 2y + 1) = - 46 + 49 + 1     where (1/2*-14)^2 =49 & (1/2*-2)^2=1
                       (x - 7)^2  +   (y - 1)^2      =   4

Considering the standard form of the equation of a circle with  center (h, k) and radius, r:

                                (x - h)^2 + (y - k)^2 = r^2 
                                (x -7)^2 + (y - k)^2 = (2)^2

Compare your equation to the standard form to get values of h, k, and r.

                                                             h=7

                                                             k=1

                                                             r=2

∴  The center is(7,1) and the radius is 2.

 
Tags: Math, Algebra
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