slove the following equations

1.) x^2 + y^2 = 4 -------(i)
x - y - k = 0 --------(ii)

From (i), we know that x=y+k
(y+k)^2+y^2-4=0
y^2+2ky+k^2+y^2-4 = 0
2y^2+2ky+(k^2-4) = 0

Because, it has one pair of solution only
Therefore, (2k)^2-4(2)(k^2-4) = 0
4k^2-8k^2+16 = 0
4k^2 = 16
k = 2 or k=-2

2.) x^2 - 4y^2 + 20 = 0 -----(a)
x + 3y - k = 0 -----(b)

Form (b), we know that x=k-3y
(k-3y)^2-4y^2+20 = 0
k^2-6y+9y^2-4y^2+20 = 0
5y^2-6y+(k^2+20) = 0

Because it has one pair of solution only
Therefore, (-6)^2-4(5)(k^2+20) = 0
36-20k^2-400 = 0
20k^2 = -364

Therefore, no real value for k as 20k^2 should be greater than or equal to 0

2006-12-16 11:27:35 補充：
(1)應該係下面:1.) x^2 + y^2 = 4 -------(i)x - y - k = 0 --------(ii)x=y+k(y+k)^2+y^2-4=0y^2+2ky+k^2+y^2-4 = 02y^2+2ky+(k^2-4) = 0(2k)^2-4(2)(k^2-4) = 04k^2-8k^2+32 = 04k^2 = 32k = 2sqrt(2) or k=-2sqrt(2)

這不是知識