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圓與直線的交點

Nov 24th 2014, 06:48

x^2+y^2+4x-2y = 0
(x+2)^2 + (y-1)^2 = 2^2+1^2 = 5
故圓C的圓心為(-2,1) , 半徑為√5

設過P之直線為 y+2 = m(x+1)
即 mx-y+m-2 = 0 .....(1式)

因為此直線與圓C相切(切線),
所以: 圓心到切線的距離,恰為圓半徑.
利用點到直線公式可得:
｜m(-2)-1+m-2｜/ √[m^2+(-1)^2] = √5
｜-m-3｜/ √(m^2+1) = √5
等式兩邊平方,等式恆成立:
(m+3)^2 / (m^2+1) = 5
(m^2+6m+9) / (m^2+1) = 5
m^2+6m+9 = 5m^2+5
4m^2 -6m -4 = 0
2m^2 -3m -2 = 0
(2m+1)(m-2) = 0
m = -1/2 , 2
代入(1式)得:
-1/2x-y-1/2-2 = 0 或 2x-y+2-2 = 0
-x-2y-1-4 = 0 或 2x-y = 0
x+2y = -5 或 2x-y = 0 .....Ans

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