What is the y-coordinate of the center of the circle that
passes through (-1,2), (3,2), and (5,4)?

The general equation of a circle is

x² + y² + Dx + Ey + F = 0

Substituting first point, (-1,2), i.e.,  x = -1, y = 2

(-1)² + (2)² + D(-1) + E(2) + F = 0
 
            1 + 4 - D + 2E + F =  0
                5 - D + 2E + F =  0
                   -D + 2E + F = -5

Substituting second point, (3,2), i.e.,  x = 3, y = 2

(3)² + (2)² + D(3) + E(2) + F = 0
 
            9 + 4 + 3D + 2E + F =  0
               13 + 3D + 2E + F =  0
                    3D + 2E + F = -13

Substituting third point, (5,4), i.e.,  x = 5, y = 4

(5)² + (4)² + D(5) + E(4) + F = 0
 
           25 + 16 + 5D + 4E + F =   0
                41 + 5D + 4E + F =   0
                     5D + 4E + F = -41

That gives you a system of three equations in 
three unknowns:

 -D + 2E + F =  -5
 3D + 2E + F = -13
 5D + 4E + F = -41

D = -2, E = -12, F = 17

The general form the equation of the circle is

                    x² + y² - 2x - 12y + 17 = 0

Now we must get it in standard form

(x - h)² + (y - k)² = r²

                    x² + y² - 2x - 12y + 17 = 0

                         x² - 2x + y² - 12y = -17
  
                     (x² - 2x) + (y² - 12y) = -17

Complete squares:
Multiply coeff's of x and y by 1/2, square the results,
add inside parentheses and also to right side.


            (x² - 2x + 1) + (y² - 12y + 36) = -17 + 1 + 36
 
Factor the trinomials in parentheses, combine numbers 
on right:

                        (x - 1)² + (y - 6)² = 20

Compare to              (x - h)² + (y - k)² = r²
                                                         
h = 1, k = 6, r² = 20, so center is (1, 6),
              __     _ 
radius = r = Ö20 = 2Ö5 

You were asked only for the y-coordinate of the center,
which is k = 6

Edwin McCravy
AnlytcPhil@aol.com