已知过点A(0,1),且斜率为k的直线l与圆c(X-2)^2+(Y-3)^2=1,相交于M,N两点(2)求证：向量AM.向量AN=定值
\u5df2\u77e5\u8fc7\u70b9A\uff080\uff0c1\uff09\u4e14\u659c\u7387\u4e3ak\u7684\u76f4\u7ebfl\u4e0e\u5706c\uff1a\uff08x-2\uff09²+\uff08y-3\uff09²=1\u76f8\u4ea4\u4e8eM\u3001N\u4e24\u70b9.\u8bbe\u76f4\u7ebfL:y=kx+1
\u7531\uff5b\uff08x-2\uff09²+\uff08y-3\uff09²=1
\uff5by=kx+1
==> (x-2) ²+\uff08kx-2\uff09²=1
==> (1+k²)x²-4(k+1)x+7=0
\u0394=16\uff08k+1)²-28(1+k²)>0
\u8bbeM(x1,y1),N(x2,y2)
\u90a3\u4e48x1+x2=4(k+1)/(k²+1),
x1x2=7/(k²+1)
\u2235\u5411\u91cfOM*\u5411\u91cfOM=12.
\u2234x1x2+y1y2=12
\u5373x1x2+(kx1+1)(kx2+1)=12
\u2234(1+k²)x1x2+k(x1+x2)-11=0
\u5373(1+k²)*7/(1+k²)+4k(k+1)/(k²+1)-11=0
\u2234(1+k²)*7+4k(k+1)-11(k²+1)=0
\u22347+4k-11=0
\u2234k=1 (\u7b26\u5408\u0394>0)
\u2234k=1
\u89e3\u7b54\u5982\u4e0b\uff1a
\u8bbe\u76f4\u7ebf\u65b9\u7a0b\u4e3ay - 1 = kx
y - kx - 1 = 0
\u5706\u5fc3\u4e3a\uff082,3\uff09\uff0c\u534a\u5f84\u4e3a1\uff0c\u6240\u4ee5\u5706\u5fc3\u5230\u76f4\u7ebf\u7684\u8ddd\u79bb\u4e3a
|3 - 2k - 1|/\u221a\uff08k² + 1\uff09
\u8981\u4f7f\u76f4\u7ebf\u548c\u5706\u6709\u4e24\u4e2a\u4ea4\u70b9
\u6240\u4ee5\u5706\u5fc3\u5230\u76f4\u7ebf\u7684\u8ddd\u79bb\u5c0f\u4e8e\u534a\u5f84
|3 - 2k - 1|/\u221a\uff08k² + 1\uff09\uff1c 1
\uff082 - 2k\uff09² \uff1c k² + 1
4k² - 8k + 4 \uff1c k² + 1
3k² - 8k + 3 \uff1c 0
\u6240\u4ee5\uff084 - \u221a7\uff09/3 \uff1c k \uff1c \uff084 + \u221a7\uff09/3
代入圆c(X-2)^2+(Y-3)^2=1
得:(x-2)^2+((kx-2)^2=1
即(1+k²)x²-(4+4k)x+7=0
需Δ=16(1+k)-28(1+k²)>0
设M(x1,y1),N(x2,y2)
则x1+x2=4(k+1)/(k²+1)
x1x2=7/(k²+1)
∴向量AM.向量AN
=(x1+y1-1)●(x2,y2-1)
=x1x2+(y1-1)(y2-1)
=x1x2+kx1*kx2
=（1+k²)x1x2
=(1+k²)*7/(1+k²)
=7
即向量AM.向量AN=定值7
法二：几何法
| AC|=2√2
过A向圆引切线AD
|AD|²=|AC|²-r²=8-1=7
根据切割线定理：
|AM||AN|=|AD|²=7
又向量AM,AN夹角为0
∴向量AM.向量AN=|AM||AN|=7
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