Question

一道高一数学题 求解

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。求t的取值范围
2。求其中面积最大的圆的方程

Answer 1

解：（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
（2）对方程进行化简
x^2+y^2-2(t+3)x+2(1-4t^2)y+16^4+9=0
[x^2-2(t+3)x+(t+3)^2]+[y^2+2(1-4t^2)y+(1-4t^2)^2]+[-(t+3)^2-(1-4t^2)^2+16t^4+9]=0
[x-(t+3)]^2+[y+(1-4t^2)]^2+[-t^2-6t-9-1-16t^4+8t^2+16t^4+9]=0
[x-(t+3)]^2+[y+(1-4t^2)]^2+[-6t-1+7t^2]=0
化简得
[x-(t+3)]^2+[y+(1-4t^2)]^2=-7t^2+6t+1=-7(t^2-6t/7+9/49)+9/7
[x-(t+3)]^2+[y+(1-4t^2)]^2=-7(t-3/7)^2+9/7
当圆面积最大时，t=3/7，半径为R=3/(根号7)
此时圆的方程为（x-24/7)^2+(y+13/49)^2=9/7
圆的最大面积为S=πR^2=9π/7