点P(4,-2)与圆x²+y²=4上任意一点边线的中点的轨迹方程是( )A、(x-2)²+(y+1)²=1 B、(x-2)²+(y+1)²=4 C、(x+4)²+(y-2)²=4 D、(x+2)²+(y-1)²=1
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![点P(4,-2)与圆x²+y²=4上任意一点边线的中点的轨迹方程是( )A、(x-2)²+(y+1)²=1 B、(x-2)²+(y+1)²=4 C、(x+4)²+(y-2)²=4 D、(x+2)²+(y-1)²=1](/uploads/image/z/3996450-18-0.jpg?t=%E7%82%B9P%EF%BC%884%2C-2%EF%BC%89%E4%B8%8E%E5%9C%86x%26%23178%3B%2By%26%23178%3B%3D4%E4%B8%8A%E4%BB%BB%E6%84%8F%E4%B8%80%E7%82%B9%E8%BE%B9%E7%BA%BF%E7%9A%84%E4%B8%AD%E7%82%B9%E7%9A%84%E8%BD%A8%E8%BF%B9%E6%96%B9%E7%A8%8B%E6%98%AF%EF%BC%88+%EF%BC%89A%E3%80%81%28x-2%29%26%23178%3B%2B%28y%2B1%29%26%23178%3B%3D1+B%E3%80%81%28x-2%29%26%23178%3B%2B%28y%2B1%29%26%23178%3B%3D4+C%E3%80%81%28x%2B4%29%26%23178%3B%2B%28y-2%29%26%23178%3B%3D4+D%E3%80%81%28x%2B2%29%26%23178%3B%2B%28y-1%29%26%23178%3B%3D1)
点P(4,-2)与圆x²+y²=4上任意一点边线的中点的轨迹方程是( )A、(x-2)²+(y+1)²=1 B、(x-2)²+(y+1)²=4 C、(x+4)²+(y-2)²=4 D、(x+2)²+(y-1)²=1
点P(4,-2)与圆x²+y²=4上任意一点边线的中点的轨迹方程是( )
A、(x-2)²+(y+1)²=1 B、(x-2)²+(y+1)²=4 C、(x+4)²+(y-2)²=4 D、(x+2)²+(y-1)²=1
点P(4,-2)与圆x²+y²=4上任意一点边线的中点的轨迹方程是( )A、(x-2)²+(y+1)²=1 B、(x-2)²+(y+1)²=4 C、(x+4)²+(y-2)²=4 D、(x+2)²+(y-1)²=1
点P(4,-2)与圆x²+y²=4上任意一点(x1,y1)连线的中点的坐标为
x=(x1+4)/2,
y=(y1-2)/2
x1=2x-4
y1=2y+2
点(x1,y1)在圆x²+y²=4上,所以有x1²+y1²=4,即
(2x-4)²+(2y+2)²=4
所求的轨迹方程为:(x-2)²+(y+1)²=1
选A、(x-2)²+(y+1)²=1
A
设中点C(x,y)..则圆上的另一点B(2x-4,2y+2)
把B带入园的方程,..化简下,,就会是A答案
选A,选择特殊点代入验证法很快的。