Question

直线Ax+By+C=0与圆x^+y^=4交于M,N两点.若满足C^=A^+B^,3Q

Answer 1

直线Ax+By+C=0---->y=kx+b,其中k=-A/B,b=-C/B,把y=kx+b,代入x2;+y2;=4,得(1+k2;)x2;+2bkx+b2;-4=0,设M(x1,y1),N(x2,y2),则 x1+x2=-2bk/(1+k2;),x1x2=(b2;-4)/(1+k2;),于是 向量OM*向量ON=x1x2+y1y2x1x2+(kx1+b)(kx2+b) =(1+k2;)x1x2+k(x1+x2)+b2; =2b2;-4+2bk2;/(1+k2;) =(2C2;-4B2;)/B2;-2A2;'BC =-2B2;C/B2;C =-2