已知过点A（0,1）的直线l，斜率为k，与圆C：（x-2）^2+(y-3)^2=1相交于M、N两个不同点。 已知过点A（0，1）的直线l，斜率为k，与圆C：（x-2）2...
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解：（1）直线l的方程为y-1=kx,即y=kx+1,带入（x-2）^2+(y-3)^2=1并整理得（k^2+1)x^2-(4+4k)x+7=0,∴△=[-（4+4k）]^2-4•7•（k^2+1)=-4(3k^2-8k+3)＜0,
解得k＜三分之（4+√7）或k>三分之（4-√7）
(2) 设M(x,y),N(m,n),
则向量OM=（x,y）,向量ON=(m,n)
且x+m=（k^2+1)分之（4+4k)，xm=（k^2+1)分之7
∵向量OM*向量ON=xm+ny=12，且ny=(km+1)(kx+1)=(k^2)mx+k(m+x)+1
∴(k^2+1)mx+k(m+x)+1=12即7+（k^2+1)分之（4k+4k^2)+1=12，解得k=1
第一问，用代入法加判别式做也可以，用点到直线的距离来做也可以。
第二问，比较简单，用切割弦定理做就可以。
(1)根据题意可以设l的方程式为y-1=kx，整理得-kx+y-1=0。
圆心坐标为(2,3)，半径=1
故，当圆心到直线距离小于1时，直线l与圆有两个不同的交点
|-2k+2|/√k^2+1<1
整理得3K^2-8k+3<0，解得 (4-√7)/2<k<(4+√7)/2
故K∈((4-√7)/2,(4+√7)/2)
(2)向量OM*向量ON=12，而OC^2=13，r^2=1
所以向量OM*向量ON=OC^2-r^2
根据切割弦定理，得出如下判断：
O、M、N三点一线，或MN为圆的直径
由于l恒过(0,1)，所以MN为圆的直径，所以直线l的斜率k=(3-1)/2=1
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