已知圆O:x^2+y^2=9,过定点P(1,2)作互相垂直的两弦,AB,CD,则线段AC的中点M的轨迹方程
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![已知圆O:x^2+y^2=9,过定点P(1,2)作互相垂直的两弦,AB,CD,则线段AC的中点M的轨迹方程](/uploads/image/z/7650224-8-4.jpg?t=%E5%B7%B2%E7%9F%A5%E5%9C%86O%3Ax%5E2%2By%5E2%3D9%2C%E8%BF%87%E5%AE%9A%E7%82%B9P%281%2C2%29%E4%BD%9C%E4%BA%92%E7%9B%B8%E5%9E%82%E7%9B%B4%E7%9A%84%E4%B8%A4%E5%BC%A6%2CAB%2CCD%2C%E5%88%99%E7%BA%BF%E6%AE%B5AC%E7%9A%84%E4%B8%AD%E7%82%B9M%E7%9A%84%E8%BD%A8%E8%BF%B9%E6%96%B9%E7%A8%8B)
已知圆O:x^2+y^2=9,过定点P(1,2)作互相垂直的两弦,AB,CD,则线段AC的中点M的轨迹方程
已知圆O:x^2+y^2=9,过定点P(1,2)作互相垂直的两弦,AB,CD,则线段AC的中点M的轨迹方程
已知圆O:x^2+y^2=9,过定点P(1,2)作互相垂直的两弦,AB,CD,则线段AC的中点M的轨迹方程
设圆上两点为A(a,b),C(c,d),M(x,y):
a² + b² = 9 (1)
c² + d² = 9 (2)
x = (a + c)/2,a + c = 2x (3)
y = (b + d)/2,b + d = 2y (4)
AP² + CP² = AC²
(a - 1)² + (b - 2)² + (c - 1)² + (d - 2)² = (a - c)² + (b - d)²
简化得:a + c + 2(b + d) - 5 = ac + bd
利用(3)(4):x + 2y - 5 = ac + bd (5)
直角三角形斜边上的中线等于斜边的一半:MP = AC/2
4MP² = AC²
4(x - 1)² + 4(y - 2)² = (a - c)² + (b - d)²
= a² - 2ac + c² + b² - 2bd + d² = (a² + b²) + (c² + d²) - 2(ac + bd)
= 9 + 9 - 2(ac + bd)
利用(5)并整理得M的轨迹:
:(x - 1/2)² + (y - 1)² = 13/4