Dynamic Geometry: P24

Let the center of the big circle be $O(0,0)$ , the origin of the $xy$ -plane. Then the center of the green circle is $Q(-9,0)$ . Let any point (or the center of the cyan circle with radius $r$ ) on the locus be $P$ . Then we note $O$ , $P$ , the tangent point of the big circle are colinear and that $OP = 25-r$ . Also $PQ = 16-r$ . Then $OP+PQ = 25-r + 16+r = 41$ for any point $P$ on the locus. Therefore the locus is an ellipse with $O$ and $Q$ being the foci.

Let the vertices of the ellipse be $A$ and $B$ , and its center $C$ . Then $A=(-25,0)$ , $B=(16,0)$ , and $C=(-4.5,0)$ . The semi-major axis $a= \dfrac {25+16}2 = 20.5$ and semi-minor axis $b = \sqrt{a^2-c^2} = \sqrt{20.5^2-4.5^2} = 20$ , where $c$ is the distance between the center and a focus. Then the area of the ellipse is $ab\pi = 20.5 \cdot 20 \pi = \boxed {410}\pi$ .

- Let us choose an arbitrary circle tangent to the green semicircle and tangent internally to the black circle. Its radius is $r$ and its center is $D(x(r);y(r))$

• We shall use the pythagorean theorem to write two equalities :

• $\begin{cases}\left(16+r\right)^2=y\left(r\right)^2+\left(x\left(r\right)-16\right)^2\\\left(25-r\right)^2=y\left(r\right)^2+\left(x\left(r\right)-25\right)^2\end{cases}$

• This allows us to find the system of parametric equation expressing the $x$ -coordinate and $y$ -coordinate in terms of $r$ :

• $\begin{cases}x\left(r\right)=\dfrac{41r}{9}\\y\left(r\right)=\dfrac{40\sqrt{\left(9-r\right)r}}{9}\end{cases}$

• Finally we can then express $y$ in terms of $x$ to get the equation of the locus: $y=\dfrac{40\sqrt{41x-x^2}}{41}$

• Now we can calculate the required area: $2\cdot \int _0^{41}\dfrac{40\sqrt{41x-x^2}}{41}=\:410\pi$

• The answer is then $410$