已知方程x2+y2-2(t+3)x+2(1-4t2)y+16t2+9=0（t属于R）的图形是圆

（1）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>0
7t^2-6t-1