13. On the coordinate plane, set $A=(-1,0), B=(1,0)$ and $P=(0, t)$ where $0 \leq t \leq 1$. As $t$ varies, $C$ is a variable point such that $P$ is the circumcentre of $\triangle A B C$. Points which are possible positions of $C$ are coloured red. Find the total area of the red region.
(2 marks)
在坐標平面上, 設 $A=(-1,0) 、 B=(1,0)$ 及 $P=(0, t)$, 其中 $0 \leq t \leq 1$ 。賞 $t$ 變化時, $C$ 是一動點, 使得 $P$ 是 $\triangle A B C$ 的外心。現把 $C$ 點的所有可能位置均塗上紅色。求紅色區域的總面積。
(2 分)
13. $\pi+2$
13. With a fixed $P(0, t)$ where $0 \leq t \leq 1$, we can construct a circle with centre $P$ and radius $P A$ (which is the same as $P B$ ). Every point on the circumference of the circle is a possible position of $C$. In particular, when $t$ takes the extreme values 0 and 1 , we get two circles with centres $(0,0)$ and $P^{\prime}=(1,1)$ respectively and with radius 1 and $\sqrt{2}$ respectively, and they provide the boundary for the red region. In the figure, the shaded region is the set of red points.

Note that $\angle A P^{\prime} B=90^{\circ}$, so sector $P^{\prime} A B$ has area $\frac{1}{4} \pi(\sqrt{2})^{2}=\frac{\pi}{2}$. Each remaining portion of the white enclosed region has area $\frac{1}{4} \pi(1)^{2}-\frac{1}{2}(1)(1)=\frac{\pi}{4}-\frac{1}{2}$, so the entire white enclosed area has area $\frac{\pi}{2}+2\left(\frac{\pi}{4}-\frac{1}{2}\right)=\pi-1$. Consequently, the total area of the shaded region $\left[\pi(\sqrt{2})^{2}-(\pi-1)\right]+\left[\pi(1)^{2}-(\pi-1)\right]=\pi+2$.