Question

在三角形ABC中,csinC=3asinA+3bsinB,作直线ax-by+c=0,被圆x^2+y^2=9所截得的弦长

Answer 1

∵a/sinA=b/sinB=c/sinC=1/k (1/k是设的,为了方便计算)

sinA=ka sinB=kb sinC=kc

csinC=3asinA+3bsinB

kc^2=3ka^2+3kb^2

c^2=3a^2+3b^2

a^2+b^2=c^2/3

圆x^2+y^2=9

圆心是(0,0),半径R=3

直线ax-by+c=0到圆心的距离

d=|a*0-b*0+c|/√(a^2+b^2)

=|c|/√(c^2/3)

=|c|/(|c|*1/√3)

=√3

∴弦长=2√(R^2-d^2)=2√(9-3)=2√6