Circles o and bc are tangent to d.

∴od⊥bc

E is eh⊥bc, bc is h,

∫∠ABC = 30

bd=√3

∴od=bdtan30 =√3*(√3/3)= 1

ob=bd/cos30 =√3/(√3/2)=2

And be = ob-OE = ob-od = 2-1=1.

∴eh=( 1/2)be= 1/2

∴s△bed=( 1/2)bd*eh=( 1/2)*√3*( 1/2)=√3/4

∫BF = ob+of = 2+x

cf⊥ab

∴fg=bf*tan30 =(x+2)*(√3/3)=(√3/3)(x+2)

∴s△bfg=( 1/2)bf*fg=( 1/2)(x+2)*(√3/3)(x+2)=(√3/6)(x+2)^2

y=s△bfg-s△bed=(√3/6)(x+2)^2-(√3/4)=(√3/ 12)(2x^2+8x+5)

That is, the functional relationship between y and x is y = (√ 3/ 12) (2x 2+8x+5).

If the area of edgf is 5 times that of △bed, it can be used.

(√3/ 12)(2x^2+8x+5)=5*(√3/4)

X 2+4x-5 = 0。

The solution is x=-5 (truncation) or x= 1.

That is = 1.

∴ef=bf-be=x+2- 1= 1+2- 1=2

And the radius of the circle o is 2*oe=2*od=2* 1=2.

You are fg⊥ab

∴ The positional relationship between the straight line where FG is located and the tangent of circle O to point F.