设m,n属于R,若直线(m+1)x+(n+1)y-2=0与圆(x-1)^2+(y-1)^2=1相切,则m+n的取值范围是
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![设m,n属于R,若直线(m+1)x+(n+1)y-2=0与圆(x-1)^2+(y-1)^2=1相切,则m+n的取值范围是](/uploads/image/z/11973330-18-0.jpg?t=%E8%AE%BEm%2Cn%E5%B1%9E%E4%BA%8ER%2C%E8%8B%A5%E7%9B%B4%E7%BA%BF%28m%2B1%29x%2B%28n%2B1%29y-2%3D0%E4%B8%8E%E5%9C%86%28x-1%29%5E2%2B%EF%BC%88y-1%29%5E2%3D1%E7%9B%B8%E5%88%87%2C%E5%88%99m%2Bn%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4%E6%98%AF)
设m,n属于R,若直线(m+1)x+(n+1)y-2=0与圆(x-1)^2+(y-1)^2=1相切,则m+n的取值范围是
设m,n属于R,若直线(m+1)x+(n+1)y-2=0与圆(x-1)^2+(y-1)^2=1相切,则m+n的取值范围是
设m,n属于R,若直线(m+1)x+(n+1)y-2=0与圆(x-1)^2+(y-1)^2=1相切,则m+n的取值范围是
因为相切 所以由距离公式得m+1+n+1/根号下(m+1)²+(n+1)²=1
化简得2mn=2(m+n)+2
因为有基本不等式 2mn≤(m+n)²/2
所以2(m+n)+2≤(m+n)²/2
所以m+n∈(-无穷,2-2根号2)∪(2+2根号2,+无穷)