4）圓通過圓C1及圓C2的交點
C1:x^2+y^2=4
C2:x^2+y^2-6x=0
已知C點點過P(2,-2),求C的方程。
A-maths!
2007-06-10 7:57 am
回答 (2)
2007-06-10 8:06 am
✔ 最佳答案
Family of circle: x^2 + y^2 - 4 + k ( x^2 + y^2 - 6x ) = 0 where k is a real constant except -1.C : x^2 + y^2 - 4 + kx^2 + ky^2 - 6kx = 0
Putting (2,-2)
(2)^2 + (-2)^2 - 4 + k(2)^2 + k(-2)^2 - 6k(2) = 0
4 + 4k + 4k - 12k = 0
4 - 4k = 0
k = 1
Therefore C: x^2 + y^2 - 4 + (1)( x^2 + y^2 - 6x ) = 0
C: x^2 + y^2 - 4 + x^2 + y^2 - 6x = 0
2x^2 + 2y^2 - 6x - 4 = 0
C: x^2 + y^2 - 3x- 2 = 0
參考: My A-maths Knowledge
2007-06-10 8:10 am
C1:x^2+y^2-4=0----------------(1),C2:x^2+y^2-6x=0------------------(2)
C1+K(C2)=0
x^2+y^2-4+k(x^2+y^2-6x)=0------------------------(3)
因為圓通過(2,-2),所以代入(3)
2^2+(-2)^2-4+k[2^2+(-2)^2-6(2)]=0
k=1--------------------(4)
把(4)代入(3):
x^2+y^2-4+x^2+y^2-6x=0
2x^2+2y^2-6x-4=0
x^2+y^2-3x-2=0
C1+K(C2)=0
x^2+y^2-4+k(x^2+y^2-6x)=0------------------------(3)
因為圓通過(2,-2),所以代入(3)
2^2+(-2)^2-4+k[2^2+(-2)^2-6(2)]=0
k=1--------------------(4)
把(4)代入(3):
x^2+y^2-4+x^2+y^2-6x=0
2x^2+2y^2-6x-4=0
x^2+y^2-3x-2=0
參考: 自己
收錄日期: 2021-04-23 20:34:15
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https://hk.answers.yahoo.com/question/index?qid=20070609000051KK05793