Dynamic Geometry: P101

We can use the property of the orthocenter that its reflection about any side of the triangle lies on the circumcircle of that triangle. No matter where the red point is, the green point's reflection about the yellow chord will always fall on the black circle. By symmetry, the green point and its reflection trace the same curves, so the purple curve is actually a circle symmetric to the black circle about the yellow chord. The diameter of the black circle is $4 + 9 = 13$ , so its area is $\dfrac{169\pi}{4}$ , meaning that $p = 169$ , $q = 4$ and $\sqrt{p} - \sqrt{q} = 11$

Let the yellow triangle inscribed by the given circle be $ABC$ , where $AB$ is the horizontal dividing chord and $C$ an arbitrary point on the circumference of the circle. Let $CD$ be the altitude from $C$ to $AB$ . Then the orthocenter $F$ of $\triangle ABC$ is colinear with $C$ and $D$ . Extend $AC$ to meet $BF$ at $G$ . Since $F$ is the orthocenter, $AG$ is perpendicular to $BF$ . Then $\angle GAB = \angle BFD$ .

Note that $ACBC'$ is a cyclic quadrilateral . Then $\angle GAB = \angle BC'D$ . Therefore $\angle BC'D = \angle BFD$ and $\triangle BDF$ and $\triangle BDC'$ are congruent since $\angle BDF = \angle BDC' = 90^\circ$ and they share $BD$ . Then every point $F$ on the locus is a reflection of point $C'$ about $AB$ . Since $C'$ traces out a circle of radius $\frac {13}2$ (see calculations in Dynamic Geometry: P96 Series ), the locus of point $F$ is also a circle of radius $\frac {13}2$ or area $\frac {13^2}{2^2}\pi$ . Therefore $\sqrt p - \sqrt q = 13 - 2 = \boxed{11}$ .