Start with two points; {A,B} and {C,D)
Then (A-x)^2 + (B-y) = R^2;
A^2 - 2Ax + x^2 + B^2 - 2By + y^2 = R^2;
(1) x^2 - 2Ax + y^2 - 2By = R^2 - A^2 - B^2; similarily
(2) x^2 - 2Cx + y^2 - 2Dy = R^2 - C^2 - D^2
(1)-(2) +2(C-A)x + 2(D-B)y = C^2 + D^2 - A^2 - B^2.
For point {E,F};
(3) x^2 - 2Ex + y^2 - 2Fy = R^2 - E^2 - F^2
(1) x^2 - 2Ax + y^2 - 2By = R^2 - A^2 - B^2; subtracting Eq(1) from Eq(3)
(3)-(1) +2(A-E)x + 2(B-F)y = A^2 + B^2 - E^2 - F^2.
Then (2)- (3) or (3)-(2) gives the third equation.
(A,B),(C,D),(E,F) are three points that are equidistant from point (x,y).
(A-x)^2 +(B-y)^2 = R^2
(C-x)^2 +(D-y)^2 = R^2
(E-x)^2 +(F-y)^2 = R^2. Then
(A-x)^2 +(B-y)^2 = R^2
A^2 + x^2 +B^2 +y^2 - 2Ax - 2Bx = R^2
(C-x)^2 +(D-y)^2 = R^2
C^2 + x^2 + D^2 +y^2 - 2Cx - 2Dx = R^2
E^2 + x^2 + F^2 +y^2 - 2Ex - 2Fx = R^2
A^2 + x^2 + B^2 + y^2 - 2Ax - 2Bx = R^2
C^2 + x^2 + D^2 + y^2 - 2Cx - 2Dx = R^2
1) A^2 - C^2 + B^2 - D^2 -2x*(A-C) - 2y*(B-D) = 0
2) A^2 - E^2 + B^2 - F^2 -2x*(A-E) - 2y*(B-F) = 0
(A+C)(A-C) + (B+C)(B-D) -2x*(A-C) - 2y*(B-D) =0
(A-C)((A+C)- 2x) + (B-D)((B+D) - 2y)
(A-E)((A+E)- 2x) + (B-F)((B+F) - 2y)
A,B
/ \ Point X,Y is the centroid of points A,B; C,D & E,F
/ \ Points G,H; I,J & K,L are points on the sides of
G,H + \ the triangle.
/ \
/ + \ If points G,H; I,J & K,L & X,Y are known,
/ X,Y \ calculate points A,B; C,D & E,F.
/ + I,J
/ \
E,F/-------------+---\ C,D
K,L
1) (A-x)^2 + (B-y) = R^2 = A^2 + B^2 + x^2 + y^2 - 2Ax - 2By = R^2
2) (C-x)^2 + (D-y) = R^2 = C^2 + D^2 + x^2 + y^2 - 2Cx - 2Dy = R^2
3) (E-x)^2 + (F-y) = R^2 = E^2 + F^2 + x^2 + y^2 - 2Ex - 2Fy = R^2
1) - 2) = 4)
4) A^2 + B^2 - C^2 - D^2 - 2Ax - 2By + 2Cx + 2Dy = 0
5) C^2 + D^2 - E^2 - F^2 - 2Cx - 2Dy + 2Ex + 2Fy = 0: 5) = 2) - 3)
6) (A-C) = (A-I)
(B-D) = (A-J)
7) (C-E) = (C-K)
(D-F) = (D-L)
b