令x-2y=k,则x=k+2y,把x=k+2y代入(x+2)²+y²=1化得：5y²+4(k+2)y+k²+4k+3=0,由此方程有实数根得：16(k+2)²-20(k²+4k+3) ≧0,解得：-2-√5≤k≤-2+√5；令（y-2)/(x-1)=m,则y-2=m(x-1),...