Dynamic Geometry: P90

Let the center of the unit semicircle be $O(0,0)$ , the origin of the $xy$ -plane, its radius be $r$ and its diameter be $AB$ , the center of the purple circle be $P$ , $\angle POB = \theta$ , and $t = \tan \frac \theta 2$ . Also let an arbitrary point on the locus or the tangent point of the center of the orange circle be $T(x,y)$ and its radius be \(r_4). From the calculations in [Dynamic Geometry: P80 Series](https://brilliant.org/discussions/thread/about-dynamic-geometry-p80-85-90-95-99-105/?ref_id=1615051), we have:

\[\begin{cases} r_4 = \dfrac {2t}{t^2+2t+9} \\ y = \dfrac {6t}{t^2+2t+9} \end{cases} \implies y = 3r_4 \]

By [Pythagorean theorem](https://brilliant.org/wiki/pythagorean-theorem/),

$\begin{aligned} x^2 + y^2 & = OT^2 = (1-r_4)^2 = \left(1-\frac y3\right)^2 \\ x^2 + \frac 89y^2 + \frac 23y & = 1 \\ x^2 + \frac 89 \left(y+\frac 38\right)^2 & = 1 + \frac 18 \\ \frac {x^2}{\frac 98} + \frac {\left(y+\frac 38\right)^2}{\frac {81}{64}} & = 1 \end{aligned}$

Therefore the locus is part of an ellipse with center at $\left(0, - \dfrac 38\right)$ , minor semi-axis of length $a = \dfrac 3{2\sqrt 2}$ and major semi-axis of length $b = \dfrac 98$ . And the area under the locus is a $2\sin^{-1} \dfrac {2\sqrt 2}3$ elliptical segment,

$A = ab \left(\sin^{-1} \frac {2\sqrt 2}3 - \frac {2\sqrt 2}9\right) = \frac {27}{16\sqrt 2} \left(\sin^{-1} \frac {2\sqrt 2}3 - \frac {2\sqrt 2}9\right) = \frac 3{2^5}\left(3^2\sqrt 2\sin^{-1} \frac {2\sqrt 2}3 - 2^2\right)$

Therefore $p+q = 3+2 = \boxed 5$ .