若集合A={(x,y)(x-1)^2+(y-a)^2=9} ,B={(x-a)^2+(y-1)^2=1},满足A∩B=Φ,则实数a的取值范围是
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![若集合A={(x,y)(x-1)^2+(y-a)^2=9} ,B={(x-a)^2+(y-1)^2=1},满足A∩B=Φ,则实数a的取值范围是](/uploads/image/z/6790551-15-1.jpg?t=%E8%8B%A5%E9%9B%86%E5%90%88A%3D%7B%EF%BC%88x%2Cy%29%28x-1%29%5E2%2B%EF%BC%88y-a%29%5E2%3D9%7D+%2CB%3D%EF%BD%9B%EF%BC%88x-a%EF%BC%89%5E2%2B%28y-1%EF%BC%89%5E2%3D1%7D%2C%E6%BB%A1%E8%B6%B3A%E2%88%A9B%3D%CE%A6%2C%E5%88%99%E5%AE%9E%E6%95%B0a%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4%E6%98%AF)
若集合A={(x,y)(x-1)^2+(y-a)^2=9} ,B={(x-a)^2+(y-1)^2=1},满足A∩B=Φ,则实数a的取值范围是
若集合A={(x,y)(x-1)^2+(y-a)^2=9} ,B={(x-a)^2+(y-1)^2=1},满足A∩B=Φ,则实数a的取值范围是
若集合A={(x,y)(x-1)^2+(y-a)^2=9} ,B={(x-a)^2+(y-1)^2=1},满足A∩B=Φ,则实数a的取值范围是
由题意得,A,B表示的圆无交点,相离
所以(a-1)^2+(a-1)^2>16 即
a2√2+1
应该是a大于1+2√2或者a小于1-2√2。可以想象成在直角坐标系上面的两个圆相离的情况,可以较轻松的解决~
注:√是根号,因为没找到标准的所以用这个对付一下