圆C1:x²+y²+2mx+4y+m²-5=0,圆C2:x²+y²-2x-2my+m²-3=0.(1)若两圆相外切,求出此时m的值及两圆外公切线长.(2)是否存在m时使两圆相交,若存在,求出m的范围,若不存在,说明理由.
人气:204 ℃ 时间:2020-01-27 09:35:23
解答
C1:(x+m)^2+(y+2)^2=3^2C2:(x-1)^2+(y-m)^2=2^2⑴、两圆外切,则两圆心间距离=5,(m+1)^2+(2+m)^2=25m^2+2 m+1+4+4m+m^2=252m^2+6m-20=0m^2+3m-10=0m=2或m=-5外公切线长=✓(5^2-1^2)=2✓6⑵、两圆相交,两...
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