已知圆x²+y²+x-6y+m=0,和直线x+2y-3=0交于p,Q两点,且op⊥oQ(o为坐标原点),求该圆的圆心坐标及半径
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![已知圆x²+y²+x-6y+m=0,和直线x+2y-3=0交于p,Q两点,且op⊥oQ(o为坐标原点),求该圆的圆心坐标及半径](/uploads/image/z/1073364-60-4.jpg?t=%E5%B7%B2%E7%9F%A5%E5%9C%86x%26%23178%3B%2By%26%23178%3B%2Bx-6y%2Bm%3D0%2C%E5%92%8C%E7%9B%B4%E7%BA%BFx%2B2y-3%3D0%E4%BA%A4%E4%BA%8Ep%2CQ%E4%B8%A4%E7%82%B9%2C%E4%B8%94op%E2%8A%A5oQ%EF%BC%88o%E4%B8%BA%E5%9D%90%E6%A0%87%E5%8E%9F%E7%82%B9%EF%BC%89%2C%E6%B1%82%E8%AF%A5%E5%9C%86%E7%9A%84%E5%9C%86%E5%BF%83%E5%9D%90%E6%A0%87%E5%8F%8A%E5%8D%8A%E5%BE%84)
已知圆x²+y²+x-6y+m=0,和直线x+2y-3=0交于p,Q两点,且op⊥oQ(o为坐标原点),求该圆的圆心坐标及半径
已知圆x²+y²+x-6y+m=0,和直线x+2y-3=0交于p,Q两点,且op⊥oQ(o为坐标原点),求该圆的圆心坐标及半径
已知圆x²+y²+x-6y+m=0,和直线x+2y-3=0交于p,Q两点,且op⊥oQ(o为坐标原点),求该圆的圆心坐标及半径
1.
由余弦定理得
a²=b²+c²-2bccosA
b²+c²-2bccosA=b²+c²+√3ab
cosA=-√3/2
A=5π/6
2.
sinA=sin(5π/6)=1/2
由正弦定理得a/sinA=b/sinB=c/sinC
b=asinB/sinA=2asinB=2√3sinB c=asinC/sinA=2asinC=2√3sinC
S+3cosBcosC
=(1/2)bcsinA +3cosBcosC
=(1/2)(2√3sinB)(2√3sinC)(1/2)+3cosBcosC
=3sinBsinC+3cosBcosC
=3cos(B-C)
当cos(B-C)=1时,S+3cosBcosC有最大值3
此时B=C=(π-5π/6)/2=π/12