Question

求过点M（2,2√3）且与圆x^2+y^2=4相切的直线的方程

Answer 1

=0时,a(x-2)+b(y-2√3)=0 转化成 a(x-2)＝0,即x-2=0

b+√3a=0时,b=－√3a,a(x-2)+b(y-2√3)=0 转化成a(x-2)－√3a(y-2√3)=0,

∵ a^2+b^2≠0,∴ a≠0,∴ a(x-2)－√3a(y-2√3)=0 转化成 (x-2)－√3(y-2√3)=0

即：x－2－√3y＋6=0,即：x-√3y+4=0