...圆的方程:X的平方+(y+1)的平方=5 直线 l:mx-y+1=m.。求救啊啊啊...

发布网友 发布时间：2024-05-07 19:07

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热心网友 时间：2024-08-16 12:42

证明：
变形直线方程 得 m（x-1）-（y-1）=0
显然不管m取什么值，当x=1，y=1时 方程为0.
直线恒过点 （1，1），斜率是m的直线。
而此点与圆心（0，-1）的距离是 根号5.
正好在圆周上。
所以你题目可能有错，m肯定和圆有交点，但也可能是相切。

热心网友 时间：2024-08-16 12:40

x-1=0,1-y=0
x=1,y=1
该直线恒过（1，1）
1^2+2^2=1+4=5
(1,1)在圆上
自己作图看看，过圆内的点的直线肯定与圆相交，一定有两个交点。

热心网友 时间：2024-08-16 12:38

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热心网友 时间：2024-08-16 12:43

把y=mx+1-m代入圆的方程，整理求二次式的判别式>0