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# Writing an equation

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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
+17705
+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
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### 3+0 Answers

#1
+17705
+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
+80815
+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

CPhill  Feb 5, 2016
edited by CPhill  Feb 5, 2016
edited by CPhill  Feb 5, 2016
#3
+80815
+10

2.  The second one is easy....

The equation  is

x^2  + y^2  = 25

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

CPhill  Feb 5, 2016

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