直线Ax+By+C=0与圆x2+y2=4交于M、N两点,若满足C2=A2+B2,则OM•ON(O为坐标原点)等于( ) A.-2 B.-1 C.0 D.1
问题描述:
直线Ax+By+C=0与圆x2+y2=4交于M、N两点,若满足C2=A2+B2,则
•OM
(O为坐标原点)等于( )ON
A. -2
B. -1
C. 0
D. 1
答
设M(x1,y1),N(x2,y2)则OM•ON=x1x2+y1y2由方程Ax+By+C=0与x2+y2=4联立消去y:(A2+B2)x2+2ACx+(C2-4B2)=0∴x1x2=C2−4B2A2+B2同理,消去x可得:y1y2=C2−4A2A2+B2∴x1x2+y1y2=2C2−4A2−4B2A2+B2又C2=A2+B...