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01-53 x2 + y2 - 8x - 6y + 21 = 0 (x - 4)2 + (y - 3)2 = 22 記圓心 = C = (4, 3) 記 P = (5, 2) CP 的斜率是 (3 - 2)/(4 - 5) = -1 因此, 所求弦的斜率 = (-1) / (-1) = 1 所求弦的方程是 (y - 2)/(x - 5) = 1 y - 2 = x - 5 x - y - 3 = 0 即 (E) 01-54 令圓的方程為: (x - r)2 + (y - r)2 = r2 x2 + y2 - 2rx - 2ry + r2 = 0 ...[1] PQ 方程是 x/3 + y/4 = 1 或 4x + 3y = 12 ...[2] 把 [2] 寫成 y = (12 - 4x)/3 代入 [1] x2 + [(12 - 4x)/3]2 - 2rx - 2r(12 - 4x)/3 + r2 = 0 9x2 + (12 - 4x)2 - 18rx - 18r(12 - 4x)/3 + 9r2 = 0 9x2 + (12 - 4x)2 - 18rx - 6r(12 - 4x) + 9r2 = 0 9x2 + 144 - 96x + 16x2 - 18rx - 72r + 24rx + 9r2 = 0 25x2 + (6r - 96)x + 144 - 72r + 9r2 = 0 Δ = 0 (6r - 96)2 - 4(25)(144 - 72r + 9r2) = 0 4(3r - 48)2 - 4(25)(144 - 72r + 9r2) = 0 (3r - 48)2 - (25)(144 - 72r + 9r2) = 0 9(r - 16)2 - (25)(9)(16 - 8r + r2) = 0 (r - 16)2 - (25)(16 - 8r + r2) = 0 (r - 16)2 - 52(4 - r)2 = 0 (r - 16)2 - 52(r - 4)2 = 0 (r - 16)2 - (5r - 20)2 = 0 [(r - 16) + (5r - 20)][(r - 16) - (5r - 20)] = 0 (6r - 36)(- 4r + 4) = 0 6((r - 6)(-4)(r - 1) = 0 r = 1 或 r = 6 (捨去, 因為 P = (3,0) ) 從切線性質可知 QR = OQ - r = 4 - 1 = 3 RP = OP - r = 3 - 1 = 2 因此 R 的座標是 (x, y) x = (3 × 3 + 2 × 0)/(3 + 2) = 9/5 y = (3 × 0 + 2 × 4)/(3 + 2) = 8/5 R = (9/5, 8/5) x = (3 × 3 + 2 × 0)/(3 + 2) = 9/5 即 (C) 02-52 從圖可知半徑 = a 圓心 = (a, 0) 方程是 (x - a)2 + y2 = a2 x2 - 2ax + y2 = 0 即 (B) 05-54 記 P 為圓心, r 為半徑, Q 為 AB 中點。 AB = 8 - 2 = 6 AQ = 6 ÷ 2 = 3 AP = r 由畢氏定理可知 PQ = √(r2 - 32) P = ( √(r2 - 32), r ) 圓方程是 [x - √(r2 - 32)]2 + (y - r)2 = r2 通過 (0, 2) 代入 x = 0, y = 2 [0 - √(r2 - 32)]2 + (2 - r)2 = r2 r2 - 32 + (2 - r)2 = r2 r2 - 9 + 4 - 4r + r2 = r2 r2 - 4r - 5 = 0 (r - 5)(r + 1) = 0 r = 5 或 r = -1 (捨去) 也可考慮通過 (0, 8) 代入 x = 0, y = 8 [0 - √(r2 - 32)]2 + (8 - r)2 = r2 r2 - 32 + (8 - r)2 = r2 r2 - 9 + 64 - 16r + r2 = r2 r2 - 16r + 55 = 0 (r - 5)(r - 11) = 0 r = 5 或 r = 11 (捨去) 圓方程是 [x - √(52 - 32)]2 + (y - 5)2 = 52 (x - 4)2 + (y - 5)2 = 52 x2 - 8x + 16 + y2 - 10y + 25 = 25 x2 + y2 - 8x - 10y + 16 = 0 即 (A) 2014-08-11 19:17:57 補充： 最後一題 05-54 可以更快。 明顯地 AB 的中點 = (0, 5) 圓心在 y = 5 上 半徑 = 5 那是最快的做法。 〔請用本方法吧。〕 2014-08-13 23:56:04 補充： 英文叫 line joining centre to the mid-point of chord ⊥ to chord. 我去查查中文，叫： 圓心至弦中點的連線垂直弦
其他解答:
從什麼地方可看出該弦是垂直於直線CP呢? ( ∴ 所求弦的斜率 = (-1) / (-1) = 1 )
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HKCEE Maths (圓) ＭＣ發問:
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最佳解答:
01-53 x2 + y2 - 8x - 6y + 21 = 0 (x - 4)2 + (y - 3)2 = 22 記圓心 = C = (4, 3) 記 P = (5, 2) CP 的斜率是 (3 - 2)/(4 - 5) = -1 因此, 所求弦的斜率 = (-1) / (-1) = 1 所求弦的方程是 (y - 2)/(x - 5) = 1 y - 2 = x - 5 x - y - 3 = 0 即 (E) 01-54 令圓的方程為: (x - r)2 + (y - r)2 = r2 x2 + y2 - 2rx - 2ry + r2 = 0 ...[1] PQ 方程是 x/3 + y/4 = 1 或 4x + 3y = 12 ...[2] 把 [2] 寫成 y = (12 - 4x)/3 代入 [1] x2 + [(12 - 4x)/3]2 - 2rx - 2r(12 - 4x)/3 + r2 = 0 9x2 + (12 - 4x)2 - 18rx - 18r(12 - 4x)/3 + 9r2 = 0 9x2 + (12 - 4x)2 - 18rx - 6r(12 - 4x) + 9r2 = 0 9x2 + 144 - 96x + 16x2 - 18rx - 72r + 24rx + 9r2 = 0 25x2 + (6r - 96)x + 144 - 72r + 9r2 = 0 Δ = 0 (6r - 96)2 - 4(25)(144 - 72r + 9r2) = 0 4(3r - 48)2 - 4(25)(144 - 72r + 9r2) = 0 (3r - 48)2 - (25)(144 - 72r + 9r2) = 0 9(r - 16)2 - (25)(9)(16 - 8r + r2) = 0 (r - 16)2 - (25)(16 - 8r + r2) = 0 (r - 16)2 - 52(4 - r)2 = 0 (r - 16)2 - 52(r - 4)2 = 0 (r - 16)2 - (5r - 20)2 = 0 [(r - 16) + (5r - 20)][(r - 16) - (5r - 20)] = 0 (6r - 36)(- 4r + 4) = 0 6((r - 6)(-4)(r - 1) = 0 r = 1 或 r = 6 (捨去, 因為 P = (3,0) ) 從切線性質可知 QR = OQ - r = 4 - 1 = 3 RP = OP - r = 3 - 1 = 2 因此 R 的座標是 (x, y) x = (3 × 3 + 2 × 0)/(3 + 2) = 9/5 y = (3 × 0 + 2 × 4)/(3 + 2) = 8/5 R = (9/5, 8/5) x = (3 × 3 + 2 × 0)/(3 + 2) = 9/5 即 (C) 02-52 從圖可知半徑 = a 圓心 = (a, 0) 方程是 (x - a)2 + y2 = a2 x2 - 2ax + y2 = 0 即 (B) 05-54 記 P 為圓心, r 為半徑, Q 為 AB 中點。 AB = 8 - 2 = 6 AQ = 6 ÷ 2 = 3 AP = r 由畢氏定理可知 PQ = √(r2 - 32) P = ( √(r2 - 32), r ) 圓方程是 [x - √(r2 - 32)]2 + (y - r)2 = r2 通過 (0, 2) 代入 x = 0, y = 2 [0 - √(r2 - 32)]2 + (2 - r)2 = r2 r2 - 32 + (2 - r)2 = r2 r2 - 9 + 4 - 4r + r2 = r2 r2 - 4r - 5 = 0 (r - 5)(r + 1) = 0 r = 5 或 r = -1 (捨去) 也可考慮通過 (0, 8) 代入 x = 0, y = 8 [0 - √(r2 - 32)]2 + (8 - r)2 = r2 r2 - 32 + (8 - r)2 = r2 r2 - 9 + 64 - 16r + r2 = r2 r2 - 16r + 55 = 0 (r - 5)(r - 11) = 0 r = 5 或 r = 11 (捨去) 圓方程是 [x - √(52 - 32)]2 + (y - 5)2 = 52 (x - 4)2 + (y - 5)2 = 52 x2 - 8x + 16 + y2 - 10y + 25 = 25 x2 + y2 - 8x - 10y + 16 = 0 即 (A) 2014-08-11 19:17:57 補充： 最後一題 05-54 可以更快。 明顯地 AB 的中點 = (0, 5) 圓心在 y = 5 上 半徑 = 5 那是最快的做法。 〔請用本方法吧。〕 2014-08-13 23:56:04 補充： 英文叫 line joining centre to the mid-point of chord ⊥ to chord. 我去查查中文，叫： 圓心至弦中點的連線垂直弦
其他解答:
從什麼地方可看出該弦是垂直於直線CP呢? ( ∴ 所求弦的斜率 = (-1) / (-1) = 1 )
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