Question

直线Ax+By+C=0与圆x^2+y^2=4相交于两点M、N,且满足C^2=A^2+B^2,则向量OM乘ON,cosji

Answer 1

圆x^2+y^2=4的圆心为原点,半径为2.

圆心到直线Ax+By+C=0的距离为d= C/√(A^2+B^2),

∵C^2=A^2+B^2,∴d=1

所以弦长|MN|=2√(|OM|²-d²)= 2√3.

由余弦定理得：cos∠MON=(4+4-12)/(2×2×2)=-1/2.

∴向量OM•ON=2×2×cos∠MON=-2.