Maths (locus)

Circle : x² + y² - 4x + 2y - 5 = 0 ...... [1]
Line : x + 3y + k = 0 ...... [2]

From [2] :
x = -3y - k ...... [3]

Put [3] into [1] :
(-3y - k )² + y² - 4(-3y - k) + 2y - 5 = 0
9y² + 6ky + k² + y² + 12y + 4k + 2y - 5 = 0
10y² + (14 + 6k)y + (k² + 4k - 5) = 0

discriminant Δ = 0
(14 + 6k)² - 4*10*(k² + 4k - 5) = 0
4k² - 8k - 396 = 0
k² - 2k - 99 = 0
(k + 9)(k - 11) = 0
k = -9 or k = 11

Let P = (a, b). x² + y² - 4x + 2y - 5 = 0(x – 2)² + (y+ 1)² = 10=> Center = (2, -1). x + 3y + k = 0y = - x/3 – k/3 => Slope = - 1/3. a² + b² - 4a + 2b - 5 = 0 … (i)a + 3b + k = 0 … (ii) (b + 1)/(a – 2)(- 1/3) = - 1 … (iii) From (iii), (b + 1)/(a – 2) = 3 b = 3a – 7 … (iv). Sub. (iv) into (i), a² + (3a– 7)² - 4a + 2(3a – 7) – 5 =0a² - 4a+ 3 = 0(a – 3)(a – 1) = 0 a = 1 or a = 3. Sub. a = 1 or a = 3 into (iv), b = - 4 or b = 2. P = (1, -4) or (3, 2). Sub. (a, b) = (1, -4) or (3, 2) into (ii), k = 11 or k = -9 I HOPE THIS CAN HELP YOU! ~ ^_^

2012-10-06 22:56:17 補充：
micatkie's method is really better them mine!