Question 1209330
<pre>
Since no one has drawn the figure, and also has not given an exact value
for the radius, I thought I would do so, with an exact solution in terms 
of trigonometric values.

We draw perpendiculars to DC at C and to EF at F, and they must intersect
at the center of the circle.  We also draw FC to make an isosceles trapezoid
from which we can get the measurements for the angles, in particular angle CFE,
which turns out to be {{{2pi/7}}} radians.  

The sum of the interior angles of a polygon with n-sides = {{{(n-2)*pi}}}
So each interior angle of the regular heptagon is {{{expr(1/7)(7-2)*pi=5pi/7}}}, so angles E and D are {{{5pi/7}}} each.

Isosceles trapezoid FEDC has sum of interior angles {{{2pi}}} so angles EFC and
FCD are each {{{expr(1/2)(2pi-2((5pi)/7))}}}{{{""=""}}}{{{2pi/7}}}
 

{{{drawing(400,400,-5,5,-5,5,

green(line(-7,-3.362481194,7,7.80214625),
line(7,-3.362481194,-7,7.80214625)),


red(circle(0,2.210832528,4.987918415)),

locate(3.3-.075,2.6+.175,B),locate(0-.075,4.2+.175,A),
locate(4.1-.075,-.8+.175,C), locate(1.8-.075,-3.8+.175,D),locate(-1.8-.075,-3.8+.175,E),
locate(-4.175,-.625,F), locate(-3.3-.075,2.6+.175,G),


line(4*cos(pi/2-0*2pi/7),4*sin(pi/2-0*2pi/7),4*cos(pi/2-1*2pi/7),4*sin(pi/2-1*2pi/7)),
line(4*cos(pi/2-1*2pi/7),4*sin(pi/2-1*2pi/7),4*cos(pi/2-2*2pi/7),4*sin(pi/2-2*2pi/7)),
line(4*cos(pi/2-2*2pi/7),4*sin(pi/2-2*2pi/7),4*cos(pi/2-3*2pi/7),4*sin(pi/2-3*2pi/7)),
line(4*cos(pi/2-3*2pi/7),4*sin(pi/2-3*2pi/7),4*cos(pi/2-4*2pi/7),4*sin(pi/2-4*2pi/7)),
line(4*cos(pi/2-4*2pi/7),4*sin(pi/2-4*2pi/7),4*cos(pi/2-5*2pi/7),4*sin(pi/2-5*2pi/7)),
line(4*cos(pi/2-5*2pi/7),4*sin(pi/2-5*2pi/7),4*cos(pi/2-6*2pi/7),4*sin(pi/2-6*2pi/7)),
line(4*cos(pi/2-6*2pi/7),4*sin(pi/2-6*2pi/7),4*cos(pi/2-7*2pi/7),4*sin(pi/2-7*2pi/7)),

locate(-.11,2.75,O),


blue(line(4cos(pi/2+4pi/7),4sin(pi/2+4pi/7),4cos(pi/2-4pi/7),4sin(pi/2-4pi/7))),
locate(-3.3,-.9,2pi/7)

)}}}

Next we'll draw in 3 perpendiculars to FC, namely EH, DJ, and OI.
Notice that angles EFH and FOI have equal measures because they are both
complements of the same angle IFO. Therefore angle FOI also measures {{{2pi/7}}}.

{{{drawing(400,400,-5,5,-5,5,







green(line(-7,-3.362481194,7,7.80214625),
line(7,-3.362481194,-7,7.80214625)),


red(circle(0,2.210832528,4.987918415)),

locate(3.3-.075,2.6+.175,B),locate(0-.075,4.2+.175,A),
locate(4.1-.075,-.8+.175,C), locate(1.8-.075,-3.8+.175,D),locate(-1.8-.075,-3.8+.175,E),
locate(-4.175,-.625,F), locate(-3.3-.075,2.6+.175,G),


line(4*cos(pi/2-0*2pi/7),4*sin(pi/2-0*2pi/7),4*cos(pi/2-1*2pi/7),4*sin(pi/2-1*2pi/7)),
line(4*cos(pi/2-1*2pi/7),4*sin(pi/2-1*2pi/7),4*cos(pi/2-2*2pi/7),4*sin(pi/2-2*2pi/7)),
line(4*cos(pi/2-2*2pi/7),4*sin(pi/2-2*2pi/7),4*cos(pi/2-3*2pi/7),4*sin(pi/2-3*2pi/7)),
line(4*cos(pi/2-3*2pi/7),4*sin(pi/2-3*2pi/7),4*cos(pi/2-4*2pi/7),4*sin(pi/2-4*2pi/7)),
line(4*cos(pi/2-4*2pi/7),4*sin(pi/2-4*2pi/7),4*cos(pi/2-5*2pi/7),4*sin(pi/2-5*2pi/7)),
line(4*cos(pi/2-5*2pi/7),4*sin(pi/2-5*2pi/7),4*cos(pi/2-6*2pi/7),4*sin(pi/2-6*2pi/7)),
line(4*cos(pi/2-6*2pi/7),4*sin(pi/2-6*2pi/7),4*cos(pi/2-7*2pi/7),4*sin(pi/2-7*2pi/7)),



locate(-.11,2.75,O),


blue(line(4cos(pi/2+4pi/7),4sin(pi/2+4pi/7),4cos(pi/2-4pi/7),4sin(pi/2-4pi/7))),
locate(-3.3,-.9,2pi/7),

blue(line(4cos(pi/2+6pi/7),4sin(pi/2+6pi/7),4cos(pi/2+6pi/7),4sin(pi/2+4pi/7)) ), 

blue(line(4cos(pi/2-6pi/7),4sin(pi/2-6pi/7),4cos(pi/2-6pi/7),4sin(pi/2-4pi/7))), 
blue(line(0,4sin(pi/2+4pi/7),0,2.2)), locate(-1.8,-.5,H),locate(-.13,-.93,I),
 locate(1.65,-.5,J), locate(-.54,1.9,2pi/7)


)}}}

Since each side of the heptagon is 1, 
(a) the hypotenuse EF of right triangle EFH is 1 and {{{FH=cos(2pi/7)}}}
(b) {{{HI=1/2}}} because HJ=ED=1 and HI is 1/2 of HJ
(c) {{{FI=FH+HI=cos(2pi/7)+1/2}}}

So from right triangle FIO, we can now find the desired radius FO

{{{(FI)/(FO)=sin(2pi/7)}}} and {{{FO=FI^""/sin(2pi/7)=(cos(2pi/7)+1/2)/sin(2pi/7)}}}

So the exact answer for the radius is

{{{radius = (cos(2pi/7)+1/2)/sin(2pi/7)}}} which can also be written as

{{{cot(2pi/7)+expr(1/2)csc(2pi/7)}}}  <--EXACT SOLUTION!

That's approximately 1.436997393, which approximately agrees with Ikleyn's,
and is even closer to Greenestamps', approximate solution. 

Edwin</pre>