Let #O and R# be the center and the radius of the big circle and #O' and r# the center and the radius of the small circle, respectively, as shown in the figure.
#=># Obviously, #R=OE=OG=2#
recall that tangent segments to a circle from an external point are equal in length, #=> EB=BF=1, CG=CF#,
#=> DeltaOBE and DeltaOBF# are congruent, and #DeltaOCF and DeltaOCG# are congruent.
Let #angleBOE=alpha, => angleBOF=alpha#,
let #angleCOF=beta, => angleCOG=beta#
#=> angleEOG=2(alpha+beta)=180^@#,
#=> alpha+beta=90^@#
#=> angleOCF=angleOCG=90-beta=alpha#,
#=> tanalpha=(BF)/(OF)=1/2#,
#=> sinalpha=1/sqrt5#
Now consider #DeltaOHO'#,
#sinalpha=(OH)/(OO')#,
#=> 1/sqrt5=(2-r)/(2+r)#
#=> sqrt5(2-r)=2+r#
#=> 2sqrt5-2=r(1+sqrt5)#
#=> r=(2sqrt5-2)/(1+sqrt5)=3-sqrt5~~0.7639# units
