题目内容
12.已知集合 M={(x,y)|y=$\sqrt{25-{x}^{2}}$,y≠0},N={(x,y)|y=-x+b},若M∩N≠∅,则实数b的取值范围是(-5,5$\sqrt{2}$].分析 由M与N,以及两集合交集不为空集,确定出b的范围即可.
解答
解:画出M与N中两函数图象,如图所示,
∵M={(x,y)|y=$\sqrt{25-{x}^{2}}$,y≠0},N={(x,y)|y=-x+b},且M∩N≠∅,
∴半圆y=$\sqrt{25-{x}^{2}}$与y=-x+b有公共点,
当直线y=-x+b与半圆相切时,圆心(0,0)到直线y=-x+b的距离d=r,即$\frac{|b|}{\sqrt{2}}$=5,
解得:b=5$\sqrt{2}$(负值舍去),
把(-5,0)代入y=-x+b得:b=-5,
则实数b的范围是(-5,5$\sqrt{2}$],
故答案为:(-5,5$\sqrt{2}$]
点评 此题考查了交集及其运算,熟练掌握交集的定义是解本题的关键.
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