Question

已知方程x^2+y^2-2(m+3)x+2(1-4m^2)y+16m^4+9=0表示一个圆

求实数m的取值范围和半径取值范围（要过程）

Answer 1

X^2+Y^2-2(m+3)X+2(1-4m^2)Y+16m^4+9=0
(X-(m+3))^2+(Y+1-4m^2)^2=(m+3)^2+(1-4m^2)^2-9-16m^4
等式左边可以化为
左=(m+3)^2+(1-4m^2)^2-9-16m^4
=m^2+6m+9+1-8m^2+16m^4-9-16m^4
=-7m^2+6m+1

1.
左=-(7m+1)(m-1)>0
=> -1/7<m<1

Answer 2

原式可化为：[x-(m+3)^2]+[y+(1-4m^2)]=-7m^2+6m+1
等号右边大于0解出m的范围
-1/7<m<1
应该是圆心的范围吧

Answer 3

m的取值范围m<-1 或m>1
半径取值范围r>1

Answer 4

题目有问题！！！！~~~~