中五附加數學(請教)

L1 : 3x - y + 3 = 0
L2 : x - 3y - 7 = 0
P : (2, -1)
設圓心(x, y)
半徑 r = | 3x - y + 3 | / √10 ... (1)
= | x - 3y - 7 | / √10 ... (2)
= √[(x - 2)^2 + (y + 1)^2] ... (3)
(1) = (2) => 9x^2 + y^2 + 9 - 6xy + 18x - 6y = x^2 + 9y^2 + 49 - 6xy - 14x + 42y
8x^2 - 8y^2 + 32x - 48y - 40 = 0
x^2 - y^2 + 4x - 6y - 5 = 0
(x + 2)^2 - (y + 3)^2 = 0
(x + 2 + y + 3)(x + 2 - y - 3) = 0
x = - y - 5 or x = y + 1
若 x = - y - 5,代入 (2) = (3),
| - y - 5 - 3y - 7 | / √10 = √[(-y - 5 - 2)^2 + (y + 1)^2]
| - 4y - 12 | / √10 = √[(-y - 7)^2 + (y + 1)^2]
16y^2 + 96y + 144 = 10(y^2 + 14y + 49 + y^2 + 2y + 1)
16y^2 + 96y + 144 = 20y^2 + 160y + 500
4y^2 + 64y + 356 = 0
y^2 + 16y + 89 = 0
判別式 16^2 - 4(89) = -100 => 無實根
若 x = y + 1 ,代入 (2) = (3),
| y + 1 - 3y - 7 | / √10 = √[(y + 1 - 2)^2 + (y + 1)^2]
4y^2 + 24y + 36 = 10(y^2 - 2y + 1 + y^2 + 2y + 1)
4y^2 + 24y + 36 = 20y^2 + 20
16y^2 - 24y - 16 = 0
8(2y + 1)(y - 2) = 0
y = 2 或 y = -1/2
若y = 2 => x = 3, 半徑 = √[(3 - 2)^2 + (2 + 1)^2] = √10
圓方程 (x - 3)^2 + (y - 2)^2 = 10
x^2 + y^2 - 6x - 4y - 3 = 0
若y = -1/2 => x = 1/2, 半徑 = √[(1/2 - 2)^2 + (-1/2 + 1)^2] = √(5/2)
圓方程 (x - 1/2)^2 + (y + 1/2)^2 = 5/2
x^2 + y^2 - x + y - 2 = 0

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