培养下一歹
幼苗
共回答了25个问题采纳率:92% 举报
此题只要点(1,2)在圆外,既满足条件.
x^2+y^2+kx+2y+10
=x^2+kx+k^2/4+y^2+2y+1+9-k^2/4
=(x+k/2)^2+(y+1)^2+9-k^2/4=0
于是有:(x+k/2)^2+(y+1)^2=k^2/4-9
代入x=1,y=2后,必有:
(1+k/2)^2+(2+1)^2>k^2/4-9 (1)
k^2/4-9>0 (2)
解(1)得:k>-19
解(2)得:k>6或k<-6
于是解为(-19,-6)并上(6,正无穷)
1年前
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