+0  
 
0
299
3
avatar

Write an equation of a circle that contains each set of points.

 

1. A(-2,2), B(2,-2), C(6,2)

 

 

2. R(5,0), S-(5,0), T- (0, -5)

Guest Feb 4, 2016

Best Answer 

 #1
avatar+17652 
+15

This can be done several ways; one way is purely by algebra:

 

The equation of a circle is x2 + y2 + Ax + By + C = 0.

 

For the point:  (-2,2)   --->    x = -2  and  y = 2

     substituting:   (-2)2 + (2)2 + A(-2) + B(2) + C  =  0

          --->               4 + 4 - 2A + 2B + C  =  0

          --->                         - 2A + 2B + C  =  -8        <---  Equation #1

 

For the point:  (2,-2)   --->    x = 2  and  y = -2

     substituting:   (2)2 + (-2)2 + A(2) + B(-2) + C  =  0

          --->               4 + 4 + 2A - 2B + C  =  0

          --->                           2A - 2B + C  =  -8     <---   Equation #2

 

Putting these two equations together:     -2A + 2B + C  =  -8

                                                                 2A - 2B + C  =  -8

Adding down:                                                          2C  =  -16     --->   C  =  -8

 

Putting this back into the two equations gives us:

     Equation #1:  -2A + 2B  =  0  --->   -A + B = 0

     Equation #1:   2A - 2B  =  0   --->    A - B  =  0

 

For the point:  (6,2)   --->    x = 6  and  y = 2     and  C = -8

     substituting:   (6)2 + (2)2 + A(6) + B(2) + C  =  0

          --->              36 + 4 + 6A + 2B + -8  =  0

          --->                                  6A + 2B  =  -32      --->  3A + B  =  -16

Putting this with the equation #2:                                       A - B  =  0

Adding down the columns:                                              4A  =  -16   --->   A  =  -4

Substituting this back into  A - B  =  0     gives  -4 - B  =  0     --->     B  =  -4

 

Placing  A = -4,  B = -4,  and  C = -8  into the equation  x2 + y2 + Ax + By + C = 0

gives  x2 + y2 - 4x - 4y - 8 = 0.

geno3141  Feb 5, 2016
Sort: 

3+0 Answers

 #1
avatar+17652 
+15
Best Answer

This can be done several ways; one way is purely by algebra:

 

The equation of a circle is x2 + y2 + Ax + By + C = 0.

 

For the point:  (-2,2)   --->    x = -2  and  y = 2

     substituting:   (-2)2 + (2)2 + A(-2) + B(2) + C  =  0

          --->               4 + 4 - 2A + 2B + C  =  0

          --->                         - 2A + 2B + C  =  -8        <---  Equation #1

 

For the point:  (2,-2)   --->    x = 2  and  y = -2

     substituting:   (2)2 + (-2)2 + A(2) + B(-2) + C  =  0

          --->               4 + 4 + 2A - 2B + C  =  0

          --->                           2A - 2B + C  =  -8     <---   Equation #2

 

Putting these two equations together:     -2A + 2B + C  =  -8

                                                                 2A - 2B + C  =  -8

Adding down:                                                          2C  =  -16     --->   C  =  -8

 

Putting this back into the two equations gives us:

     Equation #1:  -2A + 2B  =  0  --->   -A + B = 0

     Equation #1:   2A - 2B  =  0   --->    A - B  =  0

 

For the point:  (6,2)   --->    x = 6  and  y = 2     and  C = -8

     substituting:   (6)2 + (2)2 + A(6) + B(2) + C  =  0

          --->              36 + 4 + 6A + 2B + -8  =  0

          --->                                  6A + 2B  =  -32      --->  3A + B  =  -16

Putting this with the equation #2:                                       A - B  =  0

Adding down the columns:                                              4A  =  -16   --->   A  =  -4

Substituting this back into  A - B  =  0     gives  -4 - B  =  0     --->     B  =  -4

 

Placing  A = -4,  B = -4,  and  C = -8  into the equation  x2 + y2 + Ax + By + C = 0

gives  x2 + y2 - 4x - 4y - 8 = 0.

geno3141  Feb 5, 2016
 #2
avatar+78557 
+15

Thanks, geno....here's another option for the first one......since the distance from the center to all points is the same....set the radiuses equal....so, using the first and third point, we have 

 

(-2 - h)^2 + (2- k)^2  = (6 - h)^2 + (2-k)^2    subtract (2-k)^2  from both sides

 

(h + 2)^2    =  (h - 6)^2   expand

 

h^2 + 4h + 4   = h^2 -12h + 36      subtract h^2 from both sides

 

4h + 4  = -12h + 36    add 12h to both sides........subtract 4 from each side

 

16h  = 32     divide both sides by 16

 

h = 2   this is the x coordinate of the center    

                                                                     

Since the point   (2, -2) is on the graph and the x coordinate of the center  is 2.......the radius must be  [2 - (-2)] = 4

 

And using (2, -2)....we can find the y coordinate of the center = k, thusly

 

(2 - 2)^2 + (-2 - k)^2  = 16

 

k^2 + 4k  + 4  = 16

 

k^2 + 4k - 12  = 0

 

(k -2)(k + 6)  = 0      so   = -6  or k = 2     but k can't = - 6   since the lowest y coordinate on the graph is -2.....so k = 2

 

Ad the equation of the circle  is

 

(x - 2)^2  + (y - 2)^2  = 4^2

 

(x - 2)^2  + ( y - 2)^2  = 16

 

Here's the graph :  https://www.desmos.com/calculator/bggiezt02g

 

 

cool cool cool

CPhill  Feb 5, 2016
edited by CPhill  Feb 5, 2016
edited by CPhill  Feb 5, 2016
 #3
avatar+78557 
+10

2.  The second one is easy....

 

The equation  is

 

x^2  + y^2  = 25

 

See this, here: https://www.desmos.com/calculator/rn8doo57zh

 

 

cool cool cool

CPhill  Feb 5, 2016

11 Online Users

avatar
avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details