The equation of the locus of a point Q which is equidistant from the intersection of the circle x^2+y^2-4x-2y+3 = 0 and the straight line y=2x is
A. x+2y-4=0
B. x+2y-5 = 0
C. 2x+y-4 = 0
D. 2x+y-5 = 0
The eqution of Locus?
2016-03-12 3:47 pm
回答 (1)
2016-03-12 4:04 pm
✔ 最佳答案
x² + y² - 4x - 2y + 3 = 0 ...... ① [ Centre (2,1) ]2x - y = 0 ...... ②
Let L be the equation of the locus
∵ equidistant from the intersection of ① and ②
∴ L ⊥ ②
L：x + 2y - c = 0 ..... c is a constant
Sut (2,1) into x + 2y - c = 0
2 + 2(1) - c = 0
c = 4
∴ A. x + 2y - 4 = 0
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