已知圆C:x^2+y^2+kx+2k+k^2=0和定点P(1,-1),若过点P作圆的切线有两条，求k的取值范围。

(x+k/2)^2+y^2=-3k^2/4-2k
圆心C(-k/2,0),R^2=-3k^2/4-2k

两条切线则P在圆外
所以CP>R
CP^2>R^2
(-k/2-1)^2+(0+1)^2>-3k^2/4-2k
k^2/4+k+1+1>-3k^2/4-2k
k^2+3k+2>0
(k+1)(k+2)>0
k<-2,k>-1

x^2+y^2+kx+2k+k^2=0
x^2+kx+k^2/4+y^2+2k+3k^2/4=0
(x+k/2)^2+y^2=-2k-3k^2/4
圆心到定点的距离一定要大于半径才能达到题意
√[(-k/2-1)^2+(0-(-1)^2]>√(-2k-3k^2/4)
(k/2+1)^2+1>-2k-3k^2/4
k^2/4+k+1+1>-2k-3k^2/4
k^2+3k+2>0
(k+2)(k+1)>0
k>-1 或 k<-2

x^2+y^2+kx+2k+k^2=0
(x+(k/2))^2+y^2=-(3/4)k^2-2k
圆心C(-k/2,0), R^2=-(3/4)k^2-2k
PC^2>R^2
((k/2)+1)^2+1>-(3/4)k^2-2k
k^2+3k+2>0
k>-1
或k<-2