y圆:(x-1)²+(y+1)²=1;
圆心(1,-1)半径1;
设切线为k(x+3)=y;即kx-y+3k=0;
圆心到切线距离=|k+1+3k|/√(k²+1)=1;
|4k+1|/√(1+k²)=1;
(4k+1)²=1+k²;
16k²+8k+1=1+k²;
15k²+8k=0;
k=0或k=-8/15;
所以切线为y=0或-8x/15-y-8/5=0;
如果本题有什么不明白可以追问,
已知圆x^2+y^2-2x+2y+1=0,点A(-3,0),求过点A的圆的切线方程.
已知圆x^2+y^2-2x+2y+1=0,点A(-3,0),求过点A的圆的切线方程.
数学人气:583 ℃时间:2020-03-31 21:54:57
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