Question

已知方程：x方+y方-2（t+3）x+2（1-4t方）y+16t的四次方+9=0

Answer 1

1)

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=-7t^2+6t+1

R^2=-7t^2+6t+1

-7t^2+6t+1>0

7t^2-6t-1