Dynamic Geometry: P96

$\red{\text{[Updated with a simpler solution.]}}$ Let the center of the circle be $O$ ; the triangle be $ABC$ , where $AB$ is the line dividing the circle into two segments; and $DE$ be the diameter of the circle intersecting $AB$ perpendicularly at $F$ . Then $DF = 4$ , $FE=9$ , and the diameter of the circle $DE = DF+FE = 13$ and the radius of the circle is $6.5$ . Then $OF= 2.5$ and by intersecting chords theorem $AF \cdot FB = DF \cdot FE = 4 \cdot 9 \implies AF = FB = 6 \implies AB=12$ .

Let $O$ be $(0,0)$ , the origin of the $xy$ -plane, $C$ be $(u,v)$ , and an arbitrary point on the locus be $P(x,y)$ . Then $A=(-6,2.5)$ , $B=(6,2.5)$ , and $u^2 + v^2 = 6.5^2$ . Since $P(x,y)$ is the centroid of $\triangle ABC$ , then

$\begin{cases} x = \dfrac {-6 + u + 6}3 = \dfrac u3 & \implies u = 3x \\ y = \dfrac {2.5 + v + 2.5}3 = \dfrac {v+5}3 & \implies v = 3y - 5 \end{cases} \\ \begin{aligned} u^2 + v^2 & = 6.5^2 \\ \implies x^2 + \left(y - \frac 53\right)^2 & = \left(\frac {13}6 \right)^2 \end{aligned}$

Therefore the locus is a circle with center at $\left(0, \dfrac 53 \right)$ , radius $\dfrac {13}6$ and area $\dfrac {13^2}{6^2}\pi$ . Hence $\sqrt p - \sqrt 6 = 13 - 6 = \boxed 7$ .