已知根号2+1>r>0则两圆x²+y²=r²与(x-1)²+(y+1)²=2的位置关系怎么求?
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/02 12:56:45
![已知根号2+1>r>0则两圆x²+y²=r²与(x-1)²+(y+1)²=2的位置关系怎么求?](/uploads/image/z/8633502-54-2.jpg?t=%E5%B7%B2%E7%9F%A5%E6%A0%B9%E5%8F%B72%2B1%EF%BC%9Er%EF%BC%9E0%E5%88%99%E4%B8%A4%E5%9C%86x%26%23178%3B%2By%26%23178%3B%3Dr%26%23178%3B%E4%B8%8E%EF%BC%88x-1%EF%BC%89%26%23178%3B%2B%EF%BC%88y%2B1%EF%BC%89%26%23178%3B%3D2%E7%9A%84%E4%BD%8D%E7%BD%AE%E5%85%B3%E7%B3%BB%E6%80%8E%E4%B9%88%E6%B1%82%3F)
已知根号2+1>r>0则两圆x²+y²=r²与(x-1)²+(y+1)²=2的位置关系怎么求?
已知根号2+1>r>0则两圆x²+y²=r²与(x-1)²+(y+1)²=2的位置关系怎么求?
已知根号2+1>r>0则两圆x²+y²=r²与(x-1)²+(y+1)²=2的位置关系怎么求?
两圆x²+y²=r²与(x-1)²+(y+1)²=2
两圆的半径分别为r1=r和r2=√2
两个圆心之间的距离d=√2
∵√2+1