已知方程x²+y²-2(t+3)x+2(1-4t²)y+16t四次方+9=0表示一个圆.求

根据题意得
配方得：(x-t-3)^2+(y+1-4t^2)^2 =-(7t+1)(t-1)
-(7t+1)(t-1)>0
-1/7＜t＜1
配方：
[x-(t+3)]^2+[y+(1-4t^2)]^2=(t+3)^2+(1-4t^2)^2-16t^4-9
[x-(t+3)]^2+[y+(1-4t^2)]^2=-7t^2+6t+1
[x-(t+3)]^2+[y+(1-4t^2)]^2=-7(t^2-3/7)^2+16/7
因此r^2