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標題:
我將３個坐標轉成３條式...然後就沒辦法計算1. 9+3E+F=02. 26-D+5E+F=03. 20+2D+4E+F=0?
最佳解答:
(a) 設圓心為 (x,y) , 半徑為 r r^2 = x^2 + (y-3)^2 = (x+1)^2 + (y-5)^2 = (x-2)^2 + (y-4)^2 每一項皆減 x^2+y^2 , 等式恆成立: - 6y + 9 = 2x - 10y + 26 = - 4x - 8y + 20 - 6y + 9 = 2x - 10y + 26 4y = 2x + 17 - 6y + 9 = - 4x - 8y + 20 2y = - 4x + 11 4y = - 8x + 22 4y = 2x + 17 = - 8x + 22 x = 1/2 = 0.5 y = (2x+17)/4 = (1+17)/4 = 9/2 = 4.5 r^2 = x^2 + (y-3)^2 = 0.5^2 + (4.5-3)^2 = 1/4 + 9/4 = 10/4 = ( √10 / 2 )^2 故圓方程式為 ( x - 1/2 )^2 + ( y - 9/2 )^2 = ( √10 / 2 )^2 ..... Ans (b) d( (0.5,4.5) , (1,2) ) = √ ( 0.5^2 + 2.5^2 ) = √ 6.5 = √26 / 2 ≠ √10 / 2 故 (1,2) 不在此圓上 ..... Ans
其他解答:
我將３個坐標轉成３條式...然後就沒辦法計算1. 9+3E+F=02. 26-D+5E+F=03. 20+2D+4E+F=0?
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發問:最佳解答:
(a) 設圓心為 (x,y) , 半徑為 r r^2 = x^2 + (y-3)^2 = (x+1)^2 + (y-5)^2 = (x-2)^2 + (y-4)^2 每一項皆減 x^2+y^2 , 等式恆成立: - 6y + 9 = 2x - 10y + 26 = - 4x - 8y + 20 - 6y + 9 = 2x - 10y + 26 4y = 2x + 17 - 6y + 9 = - 4x - 8y + 20 2y = - 4x + 11 4y = - 8x + 22 4y = 2x + 17 = - 8x + 22 x = 1/2 = 0.5 y = (2x+17)/4 = (1+17)/4 = 9/2 = 4.5 r^2 = x^2 + (y-3)^2 = 0.5^2 + (4.5-3)^2 = 1/4 + 9/4 = 10/4 = ( √10 / 2 )^2 故圓方程式為 ( x - 1/2 )^2 + ( y - 9/2 )^2 = ( √10 / 2 )^2 ..... Ans (b) d( (0.5,4.5) , (1,2) ) = √ ( 0.5^2 + 2.5^2 ) = √ 6.5 = √26 / 2 ≠ √10 / 2 故 (1,2) 不在此圓上 ..... Ans
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