已知三角形ABC的顶点坐标分别为A（0,5），B(1-2),C(-3,-4)求三角形的外接圆的方程
AB^2 = c^2 = ( 0-1)^2 (5 2)^2 = 50；AC^2 = b^2 = (0 3)^2 (5 4)^2 = 90；BC^2 = a^2 = (1 3)^2 (-2 4)^2 = 20；cosC = ( a^2 b^2 - c^ ) / 2ab = ( 20 90 - 50 ) / 2√(90*20) = √2/2 = sinC；外接圆半径R = c/2sinC = 5√2 / (2√2/2) = 5；(0-a)^2 (5-b)^2 = a^2 b^2 - 10b 25 = 25， a^2 b^2 - 10b =0；①(1-a)^2...全部
AB^2 = c^2 = ( 0-1)^2 (5 2)^2 = 50；AC^2 = b^2 = (0 3)^2 (5 4)^2 = 90；BC^2 = a^2 = (1 3)^2 (-2 4)^2 = 20；cosC = ( a^2 b^2 - c^ ) / 2ab = ( 20 90 - 50 ) / 2√(90*20) = √2/2 = sinC；外接圆半径R = c/2sinC = 5√2 / (2√2/2) = 5；(0-a)^2 (5-b)^2 = a^2 b^2 - 10b 25 = 25， a^2 b^2 - 10b =0；①(1-a)^2 (-2-b)^2 = a^2 -2a 1 b^2 4b 4 = 25，a^2 -2a b^2 4b = 20，减①，14b - 2a = 20，7b - a = 10；②(-3-a)^ (-4-b)^2 = a^2 6a 9 b^2 8b 16 = 25，a^2 6a b^2 8b = 0，减①，6a 18b = 0，a = -3b，代入②7b 3b = 10，b = 1；a = -3b = -3；外接圆方程： ( x 3 )^2 ( y - 1 )^2 = 25 。
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