过点P(4,1)作圆x2+y2-2x+2y-2=0的切线,试求切线方程
人气:121 ℃ 时间:2020-05-13 08:20:20
解答
x2+y2-2x+2y-2=0(x-1)²+(y+1)²=4;圆心为(1,-1)半径=2;所以设切线为k=(y-1)/(x-4);即kx-y-4k+1=0;圆心到切线距离d=|k+1-4k+1|/√(k²+1)=2;(2-3k)²=4(k²+1);9k²-12k+4=4k²+4;5k...有个人答案是 24X-9Y-91=0
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