过原点的直线与圆X²+Y²-6x+5=0相交于A,B两点,求弦AB的中点M的轨迹方程
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![过原点的直线与圆X²+Y²-6x+5=0相交于A,B两点,求弦AB的中点M的轨迹方程](/uploads/image/z/2710448-8-8.jpg?t=%E8%BF%87%E5%8E%9F%E7%82%B9%E7%9A%84%E7%9B%B4%E7%BA%BF%E4%B8%8E%E5%9C%86X%26%23178%3B%2BY%26%23178%3B-6x%2B5%3D0%E7%9B%B8%E4%BA%A4%E4%BA%8EA%2CB%E4%B8%A4%E7%82%B9%2C%E6%B1%82%E5%BC%A6AB%E7%9A%84%E4%B8%AD%E7%82%B9M%E7%9A%84%E8%BD%A8%E8%BF%B9%E6%96%B9%E7%A8%8B)
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过原点的直线与圆X²+Y²-6x+5=0相交于A,B两点,求弦AB的中点M的轨迹方程
过原点的直线与圆X²+Y²-6x+5=0相交于A,B两点,求弦AB的中点M的轨迹方程
过原点的直线与圆X²+Y²-6x+5=0相交于A,B两点,求弦AB的中点M的轨迹方程
设直线为y=kx,代入圆的方程得:
x^2(1+k^2)-6x+5=0
A(x1,kx1),B(x2,kx2),x1+x2=6/(1+k^2)
记M(x,y),则有:x=(x1+x2)/2,y=kx
即(x1+x2)=2x,k=y/x,代入 x1+x2=6/(1+k^2),即得:
2x=6/(1+y^2/x^2)
化简即得M的轨迹方程:x^2+y^2-3x=0
逐行解释吧!
1)原方程化为标准型,(x-3)
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