25. The diagram shows two unshaded circles which touch each other and also touch a larger circle. Chord $P Q$ of the larger circle is a tangent to both unshaded circles. The length of $P Q$ is 6 units.
What is the area, in square units, of the shaded region?
A $3 \pi$
B $\frac{7 \pi}{2}$
C $4 \pi$
D $\frac{9 \pi}{2}$
E $5 \pi$
25. D Let the centre of the large circle and the point where the two smaller circles touch be $O$ and $R$ respectively. Let the radii of the two smaller circles be $r_{1}$ and $r_{2}$, where $r_{1} \geq r_{2}$. Then the radius of the large circle is $r_{1}+r_{2}$. Consider triangle $O P R: P R=\frac{1}{2} P Q=3 ; O P=r_{1}+r_{2}$ and $O R=r_{1}+r_{2}-2 r_{2}=r_{1}-r_{2}$.

Therefore, by Pythagoras' Theorem, $\left(r_{1}+r_{2}\right)^{2}=\left(r_{1}-r_{2}\right)^{2}+3^{2}$. So $r_{1}^{2}+2 r_{1} r_{2}+r_{2}^{2}=r_{1}^{2}-2 r_{1} r_{2}+r_{2}^{2}+9$. Hence $r_{1} r_{2}=\frac{9}{4}$.
Now the area of the shaded region is $\pi\left(r_{1}+r_{2}\right)^{2}-\pi r_{1}^{2}-\pi r_{2}^{2}=\pi\left(2 r_{1} r_{2}\right)=\pi \times 2 \times \frac{9}{4}=\frac{9 \pi}{2}$.
(Note that the listed possible solutions show that the answer is a value independent of how the diagram is configured. So a simpler method is to consider the case when $P Q$ is a diameter of the large circle and the two small circles each have diameters of length 3. It is left to the reader to check that this configuration gives the same answer as does the general solution above.)