Question 121821
You are going to have to trust an old man's memory on this one.  I'm almost certain that we had a similar problem back when I took geometry in high school and we had to prove that your triangles PQS, PTQ, and QTS were all similar right triangles.  Of course, that was 45 years ago and what I certainly don't remember is how I went about proving it.


Presuming my memory is correct, and the three triangles are, in fact, similar, you can use a simple proportion to calculate the two missing sides of QTS.


Note that the sides of PTQ are in proportion to 5:4:3 because 15/3 = 5, 12/3 = 4, and 9/3 = 3.


That means that the legs of QTS, namely segments QT and ST, are in proportion {{{9/4=x/3}}} where x is the measure of segment ST.


Solving for x:
{{{9/4=x/3}}}
{{{4x=27}}}
{{{x=27/4}}}


Now we know that the sum of the measures of PT and ST equal the measure of the circle diameter, so,


{{{d=12 + 27/4=48/4+27/4=75/4}}}


But we need the radius which is {{{d/2}}}, so {{{r=(1/2)(75/4)=75/8}}} or {{{9}}}{{{3/8}}} if you prefer.


Hope this helps,
John