Equations of circle

用戶225241

2008-06-29 7:08 pm

Given two points A(0,6) and B(8,2).

a. Find the equation of the straight line AB
b. If AB is the diameter of a circle C1, find the equation of C1
ci. Write down the equation of the family of the circles passing through the
points A and B
cii. Find the equation of the circle C2 passing through A and B and
touches the y-axis

回答 (1)

魏王將張遼

2008-06-29 7:36 pm

✔ 最佳答案

(a) Using the two-point form:
(y - 6)/(x - 0) = (2 - 6)/(8 - 0)
(y - 6)/x = -1/2
2y - 12 = -x
x + 2y - 12 = 0
(b) Mid-point of AB = (4, 4)
Length of AB = √[(0 - 8)2 + (6 - 2)2]
= 4√5
So radius of C1 = 2√5
Hence equation of C1 is:
(x - 4)2 + (y - 4)2 = 20
x2 + y2 - 8x - 8y + 12 = 0
(c) (i) The family is given by:
x2 + y2 - 8x - 8y + 12 + k(x + 2y - 12) = 0
where k is a variable.
(ii) Sub x = 0 into the family equation:
y2 - 8y + 12 + k(2y - 12) = 0
y2 + (2k - 8)y + 12(1 - k) = 0
Since it touches the y-axis, discriminant of the quadratic equation is zero:
(2k - 8)2 - 4(1)[12(1 - k)] = 0
(k - 4)2 - 12(1 - k) = 0
k2 - 8k + 16 - 12 + 12k = 0
k2 + 4k + 4 = 0
(k + 2)2 = 0
k = -2
So C2 is:
x2 + y2 - 8x - 8y + 12 - 2(x + 2y - 12) = 0
x2 + y2 - 10x - 12y + 36 = 0

參考: My Maths knowledge

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