诺圆(x-a)^2+(y-a)^2=b^2+1始终平分圆(x+1)^2+(y+1)^2=4的周长,则实数a,b应满足的条件?
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![诺圆(x-a)^2+(y-a)^2=b^2+1始终平分圆(x+1)^2+(y+1)^2=4的周长,则实数a,b应满足的条件?](/uploads/image/z/9304485-69-5.jpg?t=%E8%AF%BA%E5%9C%86%EF%BC%88x-a%29%5E2%2B%28y-a%29%5E2%3Db%5E2%2B1%E5%A7%8B%E7%BB%88%E5%B9%B3%E5%88%86%E5%9C%86%EF%BC%88x%2B1%29%5E2%2B%28y%2B1%29%5E2%3D4%E7%9A%84%E5%91%A8%E9%95%BF%2C%E5%88%99%E5%AE%9E%E6%95%B0a%2Cb%E5%BA%94%E6%BB%A1%E8%B6%B3%E7%9A%84%E6%9D%A1%E4%BB%B6%3F)
诺圆(x-a)^2+(y-a)^2=b^2+1始终平分圆(x+1)^2+(y+1)^2=4的周长,则实数a,b应满足的条件?
诺圆(x-a)^2+(y-a)^2=b^2+1始终平分圆(x+1)^2+(y+1)^2=4的周长,则实数a,b
应满足的条件?
诺圆(x-a)^2+(y-a)^2=b^2+1始终平分圆(x+1)^2+(y+1)^2=4的周长,则实数a,b应满足的条件?
由(x-a)^2+(y-a)^2=b^2+1,得x^2-2ax+a^2+y^2-2ay+a^2=b^2+1……(1)
由(x+1)^2+(y+1)^2=4,得x^2+2x+1+y^2+2y+1=4……(2)
(2)-(1),得:
(2a+2)x+(2a+2)y=2a^2-b^2+1(即公共弦方程)
代入(-1,-1)
得2a^2+4a-b^2+5=0
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