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Answer»

Lett center of bigger semicircle(radius=3) is R1 ,the next bigger one (radius=2) is R2 and the smaller one(radius=1) is R3 and complete circle one is R (assume its radius is r )

Now consider the triangle R2 R3 R and draw a line from R to R1 it acts as a cevianon this trisngle lengths of triangle

R2 R3 = 3

R2 R =2+R

R3 R=1+R

R1 R=3-R [bigger circle is normal and complete circle are same so radius of both coincide]

R2 R1=1

R3 R1=2

now applycosine rule at point R1

(2+r)2-12-(3-r)2/2*1*(3-r) = -[(1+r)2-22-(3-r)2/2*2*(3-r)]

solving this we get r= 12/14=6/7

p+q=13



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