SOLUTION: find the center of the circle using these three points (-4,4),(-7,3),(-8,2)

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Question 616919: find the center of the circle using these three points (-4,4),(-7,3),(-8,2)
Found 2 solutions by richwmiller, EdwinParker:
Answer by richwmiller(17219)   (Show Source): You can put this solution on YOUR website!
center | (-4, -1)
radius | 5
diameter | 10
area | 25 pi~~78.5398
perimeter | 10 pi~~31.4159
http://local.wasp.uwa.edu.au/~pbourke/geometry/circlefrom3/

Answer by EdwinParker(16)   (Show Source): You can put this solution on YOUR website!
(-4,4),(-7,3),(-8,2)
Substitute those three points for (x,y) into the standard 
equation for a circle:

(x-h)² + (y-k)² = r²

So we have this system of three equations in three unknowns:

(-4-h)² + (4-k)² = r²
(-7-h)² + (3-k)² = r²
(-8-h)² + (2-k)² = r²

Set the left sides of the first two equations equal:

(-4-h)² + (4-k)² = (-7-h)² + (3-k)² 

Rearrange so as to have difference of squares on each side:

(-4-h)² - (-7-h)² = (3-k)² - (4-k)²

Factor both sides as the difference of squares:

[(-4-h) - (-7-h)][(-4-h) + (-7-h)] = [(3-k) - (4-k)][(3-k) + (4-k)]

Remove the inner parentheses:

[-4 - h + 7 + h][-4 - h - 7 - h] = [3 - k - 4 + k][3 - k + 4 - k]

Simplify:

[3][-11 - 2h] = [-1][7 - 2k]

-33 -6h = -7 + 2k

-6h - 2k = 26

------------------------------------

Set the left sides of the first and third equations equal:

(-4-h)² + (4-k)² = (-8-h)² + (2-k)² 

Rearrange so as to have difference of squares on each side:

(-4-h)² - (-8-h)² = (2-k)² - (4-k)²

Factor both sides as the difference of squares:

[(-4-h) - (-8-h)][(-4-h) + (-8-h)] = [(2-k) - (4-k)][(2-k) + (4-k)]

Remove the inner parentheses:

[-4 - h + 8 + h][-4 - h - 8 - h] = [2 - k - 4 + k][2 - k + 4 - k]

Simplify:

[4][-12 - 2h] = [-2][6 - 2k]

-48 - 8h = -12 + 4k

-8h - 4k = 36

----------------------------------

Now we have the system of equations:

-6h - 2k = 26
-8h - 4k = 36

Solve that system and get (h,k) = (-4,-1)

That's the center.  That's all you asked for.

But if you had been asked for the equation, or if you
wanted to check your problem, you would also need to 
find the radius. 

Substitute (h,k) = (-4,-1) in any one of the original 
3 equations.  I'll pick the first one:

      (-4-h)² + (4-k)² = r²
(-4-(-4))² + (4-(-1))² = r²
      (-4+4)² + (4+1)² = r²
               0² + 5² = r²
                    5² = r²
                     5 = r

So the radius is 5, and so the equation of the circle is:

(x+4)² + (y+1)² = 5²

(x+4)² + (y+1)² = 25

Edwin


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