若直线x+y-1=0与圆x2+y2-tx+2ty+t+1=0相切,则实数t等于?

问题描述:

若直线x+y-1=0与圆x2+y2-tx+2ty+t+1=0相切,则实数t等于?

把y=1-x代入圆方程,△=0,解出t即可
x²+(1-x)²-tx+2t(1-x)+t+1=0
整理得到:2x²-(2+3t)x+(3t+2)=0
△=(2+3t)²-4×2(3t+2)=0
整理得:9t²+36t+20=0
(3t+2)(3t+10)=0
解得:t1=-2/3,
t2=-10/3