Dynamic Geometry: P77

Let the center of the unit semicircle be $O(0,0)$ , the origin of the $xy$ -plane, the center of the right yellow circle or an arbitrary point on the locus be $P(x,y)$ and the radius $r_1$ , and the center and radius of the green circle be $Q$ and $r$ , the foot of the vertical red line be $N$ , and $PM$ be perpendicular to the $x$ -axis.

By Pythagorean theorem ,

$\begin{aligned} MQ^2 + PM^2 & = PQ^2 \\ (NQ - NM)^2 + PM^2 & = PQ^2 \\ (r-r_1)^2 + y^2 & = (r+r_1)^2 \\ \implies y^2 & = (r+r_1)^2 - (r-r_1)^2 = 4r_1r \end{aligned}$

Also

$\begin{aligned} OM^2 + PM^2 & = OP^2 \\ (ON+NM)^2 + y^2 & = (1-r_1)^2 \\ (1-2r+r_1)^2 + 4r_1r & = 1 - 2r_1 + r_1^2 \\ (1+r_1)^2 - 4r(1+r_1)+4r^2 + 4r_1r & = 1 - 2r_1 + r_1^2 \\ 2r_1 - 4r + 4r^2 & = -2r_1 \\ \implies r_1 & = r - r^2 \\ y^2 & = 4r_1r = 4r^2(1-r) \\ \implies y & = 2r\sqrt{1-r} \end{aligned}$

Note that $x = OM = ON + NM = 1 - 2r + r_1 = 1 - 2r + r - r^2 = 1 - r -r^2$ . The area under the locus is given by:

$\begin{aligned} A & = \int_{-1}^1 y \ dx \\ & = \int_{-1}^1 2r\sqrt{1-r} \cdot \ce d (1-r-r^2) \\ & = \int_0^1 2r(1+2r)\sqrt{1-r} \ \ce d r & \small \blue{\text{Let }u^2 = 1-r \implies 2u \ \ce du = - \ce d r} \\ & = \int_0^1 4(1-u^2)(3-2u^2)u^2 \ \ce du \\ & = \int_0^1 4(3u^2- 5u^4+2u^6) \ \ce du \\ & = 4 - 4 + \frac 8 7 = \frac 87 \end{aligned}$

Therefore $p-q = 8-7 = \boxed 1$ .