实数x,y满足x2+y2-2x-2y+1=0,求y-4⼀x-2的取值范围

2025-05-21 10:22:07
推荐回答(2个)
回答1:

首先配方,将式子变成(x-1)^2+(y-1)^2=1,因为0<=(x-1)^2<=1,0<=(y-1)^2<=1,所以可以得出 1<=x<=2,1<=y<=2,然后 -1<=x-2<=0,-3<=y-4<=-2所以y-4/x-2的取值范围 0到1/2

回答2:

这是一个求斜率的问题,即在想x,y满足(x-1)2+(y-1)2=1条件下求圆上的点到点(2,4)的斜率范围则 为4/3到正无穷