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 R2R3R and draw a line from RtoR1 it acts as a cevian on this triangle lengths of triangle
R2R3=3
R2R=2+R
R3R=1+R
R1R=3−R [bigger circle is normal and complete circle are same so radius of both coincide]
R2R1=1
R3R1=2
now apply cosine rule at point R1
(2+r)2−12−(3−r)22×1×(3−r)=−[(1+r)2−22−(3−r)22×2×(3−r)]
solving this we get r=1214=67
p+q=13