B3. In the diagram below circles $C_{1}$ and $C_{2}$ have centres $O_{1}$ and $O_{2}$. The radii of the circles are respectively $r_{1}$ and $r_{2}$ with $r_{1}=3 r_{2}$. $C_{2}$ is internally tangent to $C_{1}$ at $P$. Chord $X Y$ of $C_{1}$ has length 20 , is tangent to $C_{2}$ at $Q$ and is parallel to $O_{2} O_{1}$. Determine the area of the shaded region: that is, the region inside $C_{1}$ but not $C_{2}$.
Solution: Let $M$ be the midpoint of $X Y$, then $O_{1} M \perp X Y$.
Hence $X M O_{1}$ is a right angled triangle and therefore
$$
r_{1}^{2}=r_{2}^{2}+10^{2}
$$

Clearly the area of the shaded region is $A=\pi\left(r_{1}^{2}-r_{2}^{2}\right)$, which from (1) gives us
$$
A=100 \pi \text {. }
$$
P.S. Alternatively, one can use $r_{1}=3 r_{2}$ and find the radii explicutly from (1): $9 r_{2}^{2}=r_{2}^{2}+100$, so $r_{2}=10 / \sqrt{8}$ and $r_{1}=3 r_{2}=30 \sqrt{8}$. The difference of areas then is $(900 / 8-100 / 8) \pi=100 \pi$.
Answer: $100 \pi$.