10. Points $A$ and $C$ lie on the circumference of a circle with radius $\sqrt{50} \cdot B$ is a point inside the circle such that $\angle A B C=90^{\circ}$. If $A B=6$ and $B C=2$, find the distance from $B$ to the centre of the circle.
(1 mark)
$A$ 和 $C$ 是一個半徑為 $\sqrt{50}$ 的圓的圓周上的兩點。 $B$ 是該圓內的一點, 使得 $\angle A B C=90^{\circ}$ 。若 $A B=6$ 而 $B C=2$, 求 $B$ 與該圓圓心的距離。
10. $\sqrt{26}$
10. Let $O$ be the centre of the circle and let $D$ be the midpoint of $A C$. Then $O D \perp A C$. By Pythagoras' Theorem, we have $A C=\sqrt{6^{2}+2^{2}}=2 \sqrt{10}$. This gives $A D=\sqrt{10}$ and hence
Next, we apply the compound angle formula to obtain
$$
\begin{aligned}
\cos \angle O A B & =\cos (\angle O A D-\angle B A C) \\
& =\cos \angle O A D \cos \angle B A C+\sin \angle O A D \sin \angle B A C \\
& =\frac{\sqrt{10}}{\sqrt{50}} \times \frac{6}{2 \sqrt{10}}+\frac{2 \sqrt{10}}{\sqrt{50}} \times \frac{2}{2 \sqrt{10}}=\frac{1}{\sqrt{2}}
\end{aligned}
$$

It thus follows from the cosine formula that
$$
O B=\sqrt{A O^{2}+A B^{2}-2 A O \times A B \cos \angle O A B}=\sqrt{26} .
$$