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文章數：297 |
 A-maths!!_0 |
| 不分類｜不分類 2017/08/03 00:07:12 |
標題:
此文章來自奇摩知識+如有不便請留言告知 A-maths!!發問: a証明A(6,0)和B(0,-4)兩點位於圓x^2+y^2-4x+y-12=0上。 b試求通過A和B點的圓切線之交點。 最佳解答: a 將A(6,0)代入x^2+y^2-4x+y-12=0 LHS = 6^2 + 0^2 - 4(6) + 0 - 12 = 0 =RHS ∴A(6,0)位於圓x^2+y^2-4x+y-12=0上 將B(0,-4)代入x^2+y^2-4x+y-12=0 LHS = 0^2 + 4^2 - 4(0) - 4 - 12 = 0 = RHS ∴B(0,-4)位於圓x^2+y^2-4x+y-12=0上 b 通過A點的圓切線: 6x + 0y - 4(6+x)÷2 + (0+y)÷2 - 12 = 0 12x - 24 - 4x + y -24 = 0 8x + y - 48 = 0---------------------------(1) 通過B點的圓切線: 0x + (-4)y - 4(0+x)÷2 + (-4+y)÷2 - 12 = 0 -8y - 4x -4 +y - 24 = 0 -12x + y - 28 =0-------------------------(2) (1) - (2): 20x - 20 = 0 x = 1 將 x=1代入(2): -12 + y - 28 =0 y = 40 ∴交點是(1,40) 其他解答: For part (a), you can simply substitute the coordinates of A and B into the equation of the circle. For point A, (6)^2+(0)^2-4(6)+(0)-12=0 For point B, (0)^2+(-4)^2-4(0)+(-4)-12=0 So point A and point B lie on the circle. But for part B, I don't really understand the question, so I cannot answer that... =,= |
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