✔ 最佳答案
1)設圓的方程為:
x2 + y2 + 5x﹣3y﹣4 + k(x﹣2y + 3) = 0,其中k為一常數
由於圓經過點(1,3),
將(1,3)代入圓的方程,
12 + 32 + 5(1)﹣3(3)﹣4 + k[1﹣2(3) + 3] = 0
-2k = -2
k = 1
∴所求圓的方程為:
x2 + y2 + 5x﹣3y﹣4 + 1(x﹣2y + 3) = 0
x2 + y2 + 6x﹣5y﹣1 = 0
2)設圓的方程為:
x2 + y2﹣2x + 6y + 5 + k(3x﹣y﹣1) = 0
即是x2 + y2﹣(2﹣3k)x + (6﹣k)y + 5﹣k = 0
半徑 = 5單位,
即是(1/2)√{[-(2﹣3k)]2 + (6﹣k)2﹣4(5﹣k)} = 5
√[(4﹣12k + 9k2) + (36﹣12k + k2)﹣20 + 4k] = 10
√(10k2﹣20k﹣20) = 10
兩邊 take 平方,
10k2﹣20k﹣20 = 100
10k2﹣20k﹣80 = 0
k2﹣2k﹣8 = 0
(k﹣4)(k + 2) = 0
k = 4 或 -2
∴所求圓的方程為:
當k = 4時,
x2 + y2﹣2x + 6y + 5 + 4(3x﹣y﹣1) = 0
x2 + y2 + 10x + 2y +1 = 0
當k = -2時,
x2 + y2﹣2x + 6y + 5﹣2(3x﹣y﹣1) = 0
x2 + y2﹣8x + 8y +7 = 0