已知圆C:X^2+y^2-2x-2y+1=0相切的直线L交X轴,Y轴于A,B两点,|OA|=a,|OB|=b,(a>2,

1个回答

  • 曲线C为圆:(x-1)^2+(y-1)^2 =1.圆心C(1,1),半径=1

    直线L:x/a +y/b =1,若直线L与圆相切,则:

    C(1,1)到直线L距离 =半径 =|1/a +1/b -1|/根号(1/a^2+1/b^2)

    ==> ab(ab-2a-2b-2)=0 ==> ab-2a-2b+2 =0

    ==> (a-2)(b-2)=2 ...(1)

    线段AB中点P(X,Y),X=a/2,Y=b/2

    (1) ==> (X-1)(Y-1)=1/2,(X,Y>1).此即轨迹方程

    三角形AOB面积S=ab/2

    ab-2a-2b+2 =0 ==> ab+2=2(a+b)>=4*根号(ab)

    ab>=6+4*根号2

    ==> S>=3+2*根号2

    面积的最小值 =3+2*根号2