(1) suppose the coordinate of the vertical center is M(x, y), vector AM=(x-4, y), vector BM=(x+2, y+2), vector CM=(x- 1, y-6), and vector BC=(3.

AM * BC = 0 BM * AC = 0 CM * BA = 0。 Solve the equation and get: m (20/7, 3/7).

(2) Assume that the epicentral distance is N(x, y), because the epicentral distance is the intersection of three vertical lines, and let the midpoint of BC side be A, AC side be B and AB side be C respectively. Then a (-1/2,2) b (5/2,3) c (1,-1)

Vector aN(x+ 1/2, y-2) vector bN(x-5/2, y-3) vector cN(x- 1, y+ 1), so there are:

AN * BC = 0 bN * AC = 0 cN * BA = 0。 Solve the equation to get x =114 and Y = 25/ 14.

The square of the distance from point N (114,25/14) to any point A, B and C is 3650/ 196.

So the equation of the circle is: (x-114) 2+(y-25/14) 2 = 3650/196.

Well, I did a rough calculation, and the answer is probably wrong ... but the method of solving this kind of problem with vectors is unchanged.