Question

已知方程x^2+y^2-2(t+3)+2(1-4t^2)y+16t^4+9=0表示一个圆，求t取值范围

Answer 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=0
(x-t-3)^2+(y+1-4t^2)^2=6t-7t^2
6t-7t^2＞0 得：0＜t＜6/7
所以：t取值范围0＜t＜6/7

Answer 2

x²+y²+2(1-4t²)y=2(t+3)-16t²·t²-9
x²+y²+2(1-4t²)y+(1-4t²)²=2(t+3)-16t²·t²-9+(1-4t²)²
x²+[y+(1-4t²)]² = -8t²+2t-2
则 -8t²+2t-2>0，4t²-t+1<0，无解 。题目有误

Answer 3

(1)由原方程得[x-(t 3)] 2; [y (1-4t 2;)] 2;=-7t 2; 6t 1 ①因为方程表示的图形为圆，所以-7t 2; 6t 1

Answer 4

将其化成圆心坐标式，将常数项移到右边，常数项大于零即可求到t范围。以后数学问题直接来找我，方法思维过程都提供

Answer 5

-1/7<t<1