圆(x+1)^2+(y-3)^2=10与直线x-y=0交于A,B两点,则AB弦长为?
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![圆(x+1)^2+(y-3)^2=10与直线x-y=0交于A,B两点,则AB弦长为?](/uploads/image/z/13551506-26-6.jpg?t=%E5%9C%86%EF%BC%88x%2B1%EF%BC%89%5E2%2B%EF%BC%88y-3%EF%BC%89%5E2%3D10%E4%B8%8E%E7%9B%B4%E7%BA%BFx-y%3D0%E4%BA%A4%E4%BA%8EA%2CB%E4%B8%A4%E7%82%B9%2C%E5%88%99AB%E5%BC%A6%E9%95%BF%E4%B8%BA%3F)
圆(x+1)^2+(y-3)^2=10与直线x-y=0交于A,B两点,则AB弦长为?
圆(x+1)^2+(y-3)^2=10与直线x-y=0交于A,B两点,则AB弦长为?
圆(x+1)^2+(y-3)^2=10与直线x-y=0交于A,B两点,则AB弦长为?
联立方程得 { x-y=0 (2)
由(2)有:y=x (3)
将(3)代入(1)有:x²-2x=0→x=0或x=2,对应y的值是0或2
设A(0.0),B(2,2),根据两点间距离公式有:
AB弦长=根号[(2-0)²+(2-0)²]=2(根号2)