Question

14.已知圆C:(x+1)2+(y+1)2=1,点P(x0,y0)在直线 x-y+2=0上.若圆C

上存在点Q使∠CPQ=

Answer 1

P(x0,y0)不好写·我把P点改成（m,n）这样好敲
设过P做圆C的切线，则切线与PC的夹角应大于30度，因为圆C的半径为1，所以PC<=2
C(-1,-1) (m+1)^2+(n+1)^2<=4 m-n+2=0
n=m+2 (m+1)^2+(m+3)^2<=4
m^2+2m+1+m^2+6m+9<=4
2m^2+8m+6<=0
m^2+4m+3<=0
-3<=m<=-1
m的范围是：-3<=m<=-1。