已知方程x2+y2-2(t+3)x+2(1-4t2)y+16t4+9=0表示一个圆则t的范围
人气：176 ℃ 时间：2020-04-15 18:09:37
解答
x2+y2-2(t+3)x+2(1-4t2)y+16t4+9=0[x-(t+3)]^2+[y+(1-4t^2)]^2=-16t^4-9+(t+3)^2+(1-4t^2)^2则-16t^4-9+(t+3)^2+(1-4t^2)^2〉0-16t^4-9+t^2+6t+9+1-8t^2+16t^4>0-7t^2+6t+1>07t^2-6t-1
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