標題:
F.5圓方程
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最佳解答:
圓C: x2 + y2 - 6x + 6y - 32 = 0 與 直線L: y = x + c 相切。 (a) 兩者相切，即是只有一個相交點。 考慮 {x2 + y2 - 6x + 6y - 32 = 0 {y = x + c 即 x2 + (x + c)2 - 6x + 6(x + c) - 32 = 0 x2 + x2 + 2cx + c2 - 6x + 6x + 6c - 32 = 0 2x2 + 2cx + c2 + 6c - 32 = 0只有重根。 Δ = 0 (2c)2 - 4(2)(c2 + 6c - 32) = 0 4c2 - 4(2)(c2 + 6c - 32) = 0 c2 - (2)(c2 + 6c - 32) = 0 (2)(c2 + 6c - 32) - c2 = 0 c2 + 12c - 64 = 0 (c + 16)(c - 4) = 0 c = -16 或 c = 4 (b) 若 c = -16，考慮 {x2 + y2 - 6x + 6y - 32 = 0 {y = x - 16 即 2x2 + 2(-16)x + (-16)2 + 6(-16) - 32 = 0 2x2 - 32x + 128 = 0 x2 - 16x + 64 = 0 (x - 8)2 = 0 x = 8 y = 8 - 16 = -8 交點坐標 = (8, -8) 若 c = 4，考慮 {x2 + y2 - 6x + 6y - 32 = 0 {y = x + 4 即 2x2 + 2(4)x + (4)2 + 6(4) - 32 = 0 2x2 + 8x + 8 = 0 x2 + 4x + 4 = 0 (x + 2)2 = 0 x = -2 y = -2 + 4 = 2 交點坐標 = (-2, 2)
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F.5圓方程
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1. 圓C: x^2+y^2-6x+6y-32=0與直線L: y=x+c相切。 a) 求c的值。 b) 求圓C與直線L的交點坐標。 望各位能幫忙 !! Thanks......最佳解答:
圓C: x2 + y2 - 6x + 6y - 32 = 0 與 直線L: y = x + c 相切。 (a) 兩者相切，即是只有一個相交點。 考慮 {x2 + y2 - 6x + 6y - 32 = 0 {y = x + c 即 x2 + (x + c)2 - 6x + 6(x + c) - 32 = 0 x2 + x2 + 2cx + c2 - 6x + 6x + 6c - 32 = 0 2x2 + 2cx + c2 + 6c - 32 = 0只有重根。 Δ = 0 (2c)2 - 4(2)(c2 + 6c - 32) = 0 4c2 - 4(2)(c2 + 6c - 32) = 0 c2 - (2)(c2 + 6c - 32) = 0 (2)(c2 + 6c - 32) - c2 = 0 c2 + 12c - 64 = 0 (c + 16)(c - 4) = 0 c = -16 或 c = 4 (b) 若 c = -16，考慮 {x2 + y2 - 6x + 6y - 32 = 0 {y = x - 16 即 2x2 + 2(-16)x + (-16)2 + 6(-16) - 32 = 0 2x2 - 32x + 128 = 0 x2 - 16x + 64 = 0 (x - 8)2 = 0 x = 8 y = 8 - 16 = -8 交點坐標 = (8, -8) 若 c = 4，考慮 {x2 + y2 - 6x + 6y - 32 = 0 {y = x + 4 即 2x2 + 2(4)x + (4)2 + 6(4) - 32 = 0 2x2 + 8x + 8 = 0 x2 + 4x + 4 = 0 (x + 2)2 = 0 x = -2 y = -2 + 4 = 2 交點坐標 = (-2, 2)
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