已知方程x2+y2-2(t+3)x+2(1-4t2)y+16t4+9=0,表示的图形是圆

Question

如题所述

Answer 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<0
(t-1)(7t+1)<0
-1/7<t<1追答

圆的一般方程：x²+y²+Dx+Ey+F=0
半径r²=(D²+E²-4F)/2
该题中，D=-2(t+3)，E=2(1-4t²)，F=16t^4+9
则D²+E²-4F=4(t+3)²+4(1-4t²)²-64t^4-36
=-28t²+24t+4
r²=-14t²+12t+2，当t=3/7时，r²有最大值32/7，此时圆的面积最大
所以，t=3/7时，圆的面积最大