Inscribed circles in a unit circle

Let the radius of the red circles be $r$ . Then the radius of the orange circle is $\frac 52 r$ . Let the tangent points between the blue and orange circle, the orange and the middle red circle, and the blue and the right red circle be $R$ , $S$ , and $T$ respectively. By Pythagorean theorem , we have:

$\begin{aligned} OP^2 & = OQ^2 - PQ^2 = (OT-QT)^2 - PQ^2 = (1-r)^2 -(2r)^2 = 1 - 2r - 3r^2 \\ \implies OP & = \sqrt{1-2r-3r^2} \end{aligned}$

Now note that:

$\begin{aligned} RS & = 2 \times \frac 52 r \\ RO + OP - PS & = 5r \\ 1 + \sqrt{1-2r-3r^2} - r & = 5r \\ \sqrt{1-2r-3r^2} & = 6r - 1 & \small \blue{\text{Squaring both sides}} \\ 1-2r-3r^2 & = 36r^2 - 12r + 1 \\ 39r^2 - 10 r & = 0 & \small \blue{\text{Since }r > 0} \\ \implies r & = \frac {10}{39} \end{aligned}$

Therefore $p+q = 10 + 39 = \boxed{49}$ .

Let the red circle be unit circles for a change and the orange one with diameter 5. Connect the centers of blue, middle red and right red to make a right triangle with base 2 (twice the radius), height h and hypothenuse r-1.
h² + 2² = (r - 1)²
h = √(r² - 2r - 3)
Let's extrapolate h to reach the point where blue and orange are touching internally.
5 + 1 = h + r
6 = √(r² - 2r - 3) + r
(6 - r)² = [√(r² - 2r - 3)]²
10r = 39
r = 39/10
When blue is the unit circle, the radius of red becomes 10/39. Answer is 10 + 39 = 49.