已知过点A(0,1)且斜率为k的直线l与圆c:(x-2)+(y-3)=1相交于M、N两点

\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

y = kx + 1\u4ee3\u5165\u5706\u7684\u65b9\u7a0b
\uff08x - 2\uff09² + \uff08kx - 2\uff09² = 1
\uff081 + k²\uff09x² - \uff084 + 4k\uff09x + 7 = 0
\u56e0\u4e3a\u4ea4\u70b9\u4e3aM\u548cN\uff0c\u6240\u4ee5\u89e3\u51fa\u6765\u7684\u4e24\u4e2ax\u4e3aM\u548cN\u7684\u6a2a\u5750\u6807
\u8bb0\u4e3ax1\u548cx2
\u6839\u636e\u97e6\u8fbe\u5b9a\u7406\u6709\uff0cx1 + x2 = \uff084 + 4k\uff09/\uff081 + k²\uff09
x1 x2 = 7/\uff081 + k²\uff09
OM * ON = \uff08x1\uff0cy1\uff09\uff08x2\uff0cy2\uff09
= x1x2 + y1y2
= 7/\uff081 + k²\uff09+ \uff08kx1 + 1\uff09\uff08kx2 + 1\uff09
= 7/\uff081 + k²\uff09+ k² * 7/\uff081 + k²\uff09+ k\uff084 + 4k\uff09/\uff081 + k²\uff09+ 1
= 8 + k\uff084 + 4k\uff09/\uff081 + k²\uff09= 12
\u6240\u4ee5k\uff084 + 4k\uff09/\uff081 + k²\uff09= 4
4k + 4k² = 4 + 4k²
\u89e3\u5f97k = 1

(1)由题意得:L: y=kx+1,代入圆的方程并整理得一元二次方程 (k+1)x2-4(k+1)x+7=0,此方程有两个不相等的实数根,所以[-4(K+1)]^2-4*7(K+1)>0,解得k<-1或k>3/4, (2)因M,N共线,向量AM,AN的夹角=0,AM*AN=|AM|*|AN|*cos0=|AM|*|AN|,过A作圆的切线AT,T为切点, 则|AM|*|AN|=|AT|^2,又A为定点,圆为定圆,所以AT的长是定值,所以向量AM与AN的内积是定值。 (3) 因O(2,3),又|AT|^2=12,半径r=1,|OA|^2=|AT|^2+R^2,所以得:|OA|^2=12+1 好困呀,咱们明天再见。

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