已知圆c x^2+(y-1)^2=5 过点(1,2)的直线交于圆A,B两点,求A,B的中点的轨迹方程
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已知圆c x^2+(y-1)^2=5 过点(1,2)的直线交于圆A,B两点,求A,B的中点的轨迹方程
已知圆c x^2+(y-1)^2=5 过点(1,2)的直线交于圆A,B两点,求A,B的中点的轨迹方程
已知圆c x^2+(y-1)^2=5 过点(1,2)的直线交于圆A,B两点,求A,B的中点的轨迹方程
设A(x1,y1)B(x2,y2)AB中点M(x,y)
2x=x1+x2,2y=y1+y2
x1²+(y1-1)²=5
x2²+(y2-1)²=5
两式相减
(x1²-x2²)+(y1-1)²-(y2-1)²=0
(x1+x2)(x1-x2)+(y1+y2-2)(y1-y2)=0
2x+(2y-2)(y1-y2)/(x1-x2)=0
因为(y1-y2)/(x1-x2)=(y-2)/(x-1)
所以2x+(2y-2)(y-2)/(x-1)=0
x(x-1)+(y-1)(y-2)=0
x²-x+y²-3y+2=0
(x-1/2)²+(y-3/2)²=1/2此为轨迹方程