I wrote:Perhaps next time I will post on how I searched upon all Pythagorean triples within the range 0..49 to find the only 2 possible combinations which satisfy all the constraints. (It isn't that hard, so I don't consider it as brute force.)

Here it is. Remember, from the

general configuration, we know that the constraints are like these:

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(r2-r1)^2 + a^2 = c^2

(r1+r2)^2 + b^2 = c^2

a, b, c are positive integers

r1, r2 are positive prime integers

c < 50

r1 < r2

a, b are different (which is implied from the fact that r2-r1 < r1+r2)---------------------------------------------------------------------------------

Firstly, we gather all the Pythagorean triples with the hypotenuse ranging up to 49. There aren't many, just 18:

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3,4,5 ; 6,8,10 ; 9,12,15 ; 12,16,20 ; 15,20,25; 18,24,30

21,28,35 ; 24,32,40 ; 27,36,45 ; 5,12,13 ; 10,24,26 ; 15,36,39

8,15,17 ; 16,30,34 ; 7,24,25 ; 20,21,29 ; 12,35,37 ; 9,40,41---------------------------------------------------------------------------------

From here on in just analyse as below:

Triple click below to see what I wrote:All these triples (x,y,z) satisfy the equation x^2 + y^2 = z^2, so we just need to try each of them to see if they can be (r2-r1,a,c) or (r1+r2,b,c).

Now it's easy to see that both r2-r1 and r1+r2 must be of the same parity (odd or even). Therefore for z to be c we need 2 different x-values or y-values with the same parity. Hence we can eliminate all the triples which only has 1 odd x-value and 1 even y-value (or vice versa) for the z-value.

So we are left with only 8 triples (with 7 different z-values):

z = 10 : x,y = 6,8

z = 20 : x,y = 12,16

z = 25 : x,y = 7,24 or 15,20

z = 26 : x,y = 10,24

z = 30 : x,y = 18,24

z = 34 : x,y = 16,30

z = 40 : x,y = 24,32

For each of these z-values, we take 2 from the x-values and y-values (with the same parity) and assign them as r2-r1 and r1+r2, and work out r1 & r2 to see if they are both prime:

z = 10 : r2-r1=6, r1+r2=8 => r1=1, r2=7

z = 20 : r2-r1=12, r1+r2=16 => r1=2, r2=14

z = 25 : r2-r1=7, r1+r2=15 => r1=4, r2=11

z = 25 : r2-r1=20, r1+r2=24 => r1=2, r2=22

z = 26 : r2-r1=10, r1+r2=24 => r1=7, r2=17

z = 30 : r2-r1=18, r1+r2=24 => r1=3, r2=21

z = 34 : r2-r1=16, r1+r2=30 => r1=7, r2=23

z = 40 : r2-r1=24, r1+r2=32 => r1=4, r2=28

From these 8 triples, the only cases which have both r1 & r2 prime are r1=7, r2=17 and r1=7, r2=23. Hence we know the radius of the small circle must be 7. (Q.E.D.)

(BTW, if we drop the constraints that r1,r2 must be prime, these 8 sets of values are the only possible answers for c<50.)