Question

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...

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

Answer 1

简单计算一下，答案如图所示

Answer 2

直线l:y=kx+1
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