Question 349269
{{{drawing(300,300,-10,10,-10,10,circle(0,0,10),
line(-6,-8,0,0),green(line(-6,-8,0,-8)),line(-6,-8,-6,-10),
line(-6,-10,-8,-10),
locate(-5,-4,R),
locate(1,-5,R),
line(-6,-8,-8,-8),
line(-8,-8,-8,-10),
blue(line(10,-10,-6,-10)),
locate(-4,-8.7,x),
locate(5,-8.7,R),
locate(-6.5,-8.5,1),
line(0,0,10,0),line(10,-10,10,10),line(0,0,0,-10))}}}
You can find the triangle with the green leg {{{x}}}since the triangle is a right triangle with the other leg {{{R-1}}} and hypotenuse of {{{R}}}.
{{{R^2=(R-1)^2+x^2}}}
.
.
{{{x+R=5}}}
{{{x=5-R}}}
{{{x^2=25-10R+R^2}}}
.
.
Substituting,
{{{R^2=R^2-2R+1+(25-10R+R^2)}}}
{{{-2R+1+25-10R+R^2=0}}}
{{{R^2-12R+26=0}}}
Complete the square,
{{{R^2-12R+36-10=0}}}
{{{(R-6)^2=10}}}
{{{R-6=0 +- sqrt(10)}}}

{{{R=6 +- sqrt(10)}}}
Since {{{x+R=5}}}, then {{{R}}} must be less than {{{5}}}. 
So, 
{{{highlight(R=6- sqrt(10))}}}