设L1斜率为k,则方程为 y=k(x-1).
联立圆C和L1方程,整理,(1+k^2)x^2-[2k(k+4)+6]x+(k+4)^2+5=0.
P(x1,y1),Q(x2,y2).
=>M((x1+x2)/2,(y1+y2)/2)或M((x1+x2)/2,k(x1+x2)/2-k).
韦达定理,
x1+x2=[2k(k+4)+6]/(1+k^2), x1x2=[(k+4)^2+5]/(1+k^2). ---(1)
联立L1与L2得 N(2(k-1)/(2k+1),-3k/(2k+1)),
验算(AM*AN)^2与k无关即可.
AM^2=[(x1+x2)/2-1]^2+[k(x1+x2)/2-k]^2,
AN^2=[2(k-1)/(2k+1)-1]^2+[-3k/(2k+1)]^2, ----(2)
由(1)(2)可得,(AM*AN)^2=9. 得证