SOLUTION: Find the centre and circumradius of a triangle with vertices A(4,3),B(-3,2) and C(1,-6).

  

Let the coordinates of the circumcenter (or "circumcentre", 
as you spell it in the UK) be O(h,k).  Then the circumradius
r = AO = BO = CO.

We use the distance formula d = 

AO = 
BO =  = 
CO =  = 

 =  = 

Squaring all three expressions:

(h-4)² + (k-3)² = (h+3)² + (k-2)² = (h-1)² + (k+6)² 

Equating the first two:

(h-4)² + (k-3)² = (h+3)² + (k-2)²

Rearrange terms to create the difference of squares:

(h-4)² - (h+3)² = (k-2)² - (k-3)² 

Factor (or as they say in UK, "factorise")

[(h-4) - (h+3)][(h-4) + (h+3)] = [(k-2) - (k-3)][(k-2) + (k-3)]

[h - 4 - h - 3][h - 4 + h + 3] = [k - 2 - k + 3][k - 2 + k - 3]

[-7][2h-1] = [1][2k-5]
-14h + 7 = 2k - 5
-14h = 2k - 12
 -7h = k - 6

Equating the lst and 3rd

(h-4)² + (k-3)² = (h-1)² + (k+6)²

Rearrange terms to create the difference of squares:

(h-4)² - (h-1)² = (k+6)² - (k-3)² 

Factor ("factorise")

[(h-4) - (h-1)][(h-4) + (h-1)] = [(k+6) - (k-3)][(k+6) + (k-3)]

[h - 4 - h + 1][h - 4 + h - 1] = [k + 6 - k + 3][k + 6 + k - 3]

[-3][2h-5] = [9][2k+3]
-6h + 15 = 18k + 27
-6h = 18k + 12
  h = -3k - 2
  
So we solve this system of two equations:

 -7h = k - 6
   h = -3k - 2

by substitution and get 

-7(-3k - 2) = k - 6
   21k + 14 = k - 6
        20k = -20
          k = -1

   h = -3(-1) - 2
   h = 3 - 2
   h = 1

So the circumcenter (or "circumcentre") is O(h,k) = O(1,-1)

The circumradius r is

r = AO =  =  =  =  =  = 5

Edwin