帮忙解一道关于圆的数学题

x^2+y^2-2(t+3)x+2(1-4t^2)y+16t^4+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>0
7t^2-6t-1<0
(t-1)(7t+1)<0
-1/7<t<1

圆的半径R满足
R^2=-7t^2+6t+1=-(t-1)(7t+1)
因为R^2>0
-1/7<t<1
而且R^2=-7(t-3/7)^2+16/7
所以该圆半径r的取值范围是(0,4/√7].