Question

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

Answer 1

根据x+y-1=0,y=1-x,代入x^2+y^2-tx+2ty+t+1=0得：

x^2+（1-x)^2-tx+2t(1-x)+t+1=0

x^2+1-2x+x^2-tx+2t-2tx+t+1=0

2x^2 - (3t+2)x + 3t+2 = 0

相切,相当于有一个交点

判别式=0

(3t+2)^2 - 4*2（3t+2)=0

(3t+2)(3t+2-8)=0

3(3t+2)(t-2)=0

t1=-2/3,t2=2