Question 388780
<pre>

The other tutor's solution is incorrect.

{{{drawing(400,400,-1.5,1.5, -1.5,1.5, 

circle(0,0,1), triangle(0,1,.67010309,-.742268,-.67010309,-.742268),
locate(.4,.1,13), locate(-.52,.1,13), locate(0,-.75,5),
locate(-.03,1.13,A),locate(-.74,-.74,B), locate(.72,-.72,C)

 )}}}

Draw in the radii to the vertices:

{{{drawing(400,400,-1.5,1.5, -1.5,1.5, 
locate(0.05,.1,O),
circle(0,0,1), triangle(0,1,.67010309,-.742268,-.67010309,-.742268),
locate(.4,.1,13), locate(-.52,.1,13), locate(0,-.75,5),
green(line(0,0,0,1), line(-.67010309,-.742268,0,0), line(.67010309,-.742268,0,0)),
locate(-.03,1.13,A),locate(-.74,-.74,B), locate(.72,-.72,C),
locate(.03,.5,r), locate(-.25,-.3,r), locate(.23,-.3,r)

 )}}} 

Extend the vertical radius down to the base of the given triangle,
which we label D, and we let x be the measure of OD
dividing the base into two equal parts, each measuring 2.5:

{{{drawing(400,400,-1.5,1.5, -1.5,1.5, 
locate(0.05,.1,O),
circle(0,0,1), triangle(0,1,.67010309,-.742268,-.67010309,-.742268),
locate(.4,.1,13), locate(-.52,.1,13), locate(0,-.75,D),
green(line(0,0,0,1), line(-.67010309,-.742268,0,0), line(.67010309,-.742268,0,0),line(0,0,0,-.742268)), 
locate(-.03,1.13,A),locate(-.74,-.74,B), locate(.72,-.72,C),
locate(.03,.5,r), locate(-.25,-.3,r), locate(.23,-.3,r),
locate(-.4,-.75,2.5), locate(.03,-.35,x)
 )}}}

From right triangle BOD  

{{{BD^2+DO^2=BO^2}}}

{{{r^2=2.5^2+x^2}}}

{{{r^2=6.25+x^2}}}

{{{r^2-x^2=6.25}}}

{{{(r-x)(r+x)=6.25}}}

From right triangle BAD

{{{(r+x)^2 + 2.5^2 = 13^2}}}

{{{(r+x)^2 + 6.25 = 169}}}

{{{(r+x)^2 + 6.25 = 169}}}

{{{(r+x)^2 = 162.75}}}

{{{r+x = "" +- sqrt(162.75)}}}

Ignore the negative:

{{{r+x =  sqrt(162.75)}}}

{{{r+x =  sqrt(162.75)}}}

{{{r+x = 12.75735082}}}


Substitute {{{sqrt(162.75)}}} for {{{r+x)}}} in

{{{(r-x)(r+x)=6.25}}}

{{{(r-x)sqrt(162.75)=6.25}}}

{{{r-x=6.25/sqrt(162.75)}}}

{{{r-x=0.489913626}}}

Add these two equations:

{{{system{r+x = 12.75735082, r-x=0.489913626}}}

2r = 13.24728445

 r = 6.62363632224

Edwin</pre>