点M(3,1),直线ax-y+4=0及圆(x-1)^2+(y-2)^2=41求过点M的圆的切线方程2若直线ax-y+4=0与圆相切,求a
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![点M(3,1),直线ax-y+4=0及圆(x-1)^2+(y-2)^2=41求过点M的圆的切线方程2若直线ax-y+4=0与圆相切,求a](/uploads/image/z/12562060-4-0.jpg?t=%E7%82%B9M%283%2C1%29%2C%E7%9B%B4%E7%BA%BFax-y%2B4%3D0%E5%8F%8A%E5%9C%86%28x-1%29%5E2%2B%28y-2%29%5E2%3D41%E6%B1%82%E8%BF%87%E7%82%B9M%E7%9A%84%E5%9C%86%E7%9A%84%E5%88%87%E7%BA%BF%E6%96%B9%E7%A8%8B2%E8%8B%A5%E7%9B%B4%E7%BA%BFax-y%2B4%3D0%E4%B8%8E%E5%9C%86%E7%9B%B8%E5%88%87%EF%BC%8C%E6%B1%82a)
点M(3,1),直线ax-y+4=0及圆(x-1)^2+(y-2)^2=41求过点M的圆的切线方程2若直线ax-y+4=0与圆相切,求a
点M(3,1),直线ax-y+4=0及圆(x-1)^2+(y-2)^2=4
1求过点M的圆的切线方程
2若直线ax-y+4=0与圆相切,求a
点M(3,1),直线ax-y+4=0及圆(x-1)^2+(y-2)^2=41求过点M的圆的切线方程2若直线ax-y+4=0与圆相切,求a
1,圆(x-1)^2+(y-2)^2=4 的圆心O为(1,2), 半径为R=2.
过点M(3,1)的直线L与圆相切,|MO|^2=(3-1)^2+(1-2)^2=5,
设切线的斜率为k,则:R^2+(kR)^2=|MO|^2,
即 4+4k^2=5,解得:k=1/4,或k=-1/4.
故所求过点M的圆的切线方程为:y-1=1/4(x-3),或 y-1=-1/4(x-3),
即 x-4y+1=0,或 x+4y-7=0.
2,直线ax-y+4=0与圆(x-1)^2+(y-2)^2=4相切,
将y=ax+4代入(x-1)^2+(y-2)^2=4,化简得:
(a^2+1)x^2+(4a-2)x+1=0,
由于直线与圆相切,上方程只有一个解.
所以 (4a-2)^2-4(a^2+1)=0,解得: a=0,或 a=4/3.
(1)由题意可知M在圆(x-1)2+(y-2)2=4外,
故当x= 3时满足与圆相切.
当斜率存在时设为y-1=k(x-3),即kx-y-3k+1=0.
由|k-2+1-3k|k2+1=2,∴k=3/4,
∴所求的切线方程为x=3或3x-4y-5=0.
(2)由ax-y+4=0与圆相切知|a-2+4|1+a2=2,
∴a=0或a=4/3.
(...
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(1)由题意可知M在圆(x-1)2+(y-2)2=4外,
故当x= 3时满足与圆相切.
当斜率存在时设为y-1=k(x-3),即kx-y-3k+1=0.
由|k-2+1-3k|k2+1=2,∴k=3/4,
∴所求的切线方程为x=3或3x-4y-5=0.
(2)由ax-y+4=0与圆相切知|a-2+4|1+a2=2,
∴a=0或a=4/3.
(3)圆心到直线的距离d=|a+2|1+a2,
又l=23,r=2,
∴由r2=d2+(l2)2,可得a=-3/4.
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