9. On the rectangular coordinate plane, a point $P$ is randomly chosen on the line segment joining $(0,-1)$ and $(0,1)$. Another point $Q$ is randomly chosen on the line segment joining $(1,-1)$ and $(1,1)$. Two circles, each with radius 1 , are then drawn using $P$ and $Q$ as centres respectively. Find the probability that the two circles intersect.
在直角坐標平面上, 我們隨意在連接 $(0,-1)$ 和 $(0,1)$ 的線段上選一點 $P$, 再隨意在連接 $(1,-1)$ 和 $(1,1)$ 的線段上選一點 $Q$, 然後分別以 $P$ 和 $Q$ 為圓心，各作一個半徑為 1 的圓。求兩圓相交的概率。
9. $\frac{4 \sqrt{3}-3}{4}$
9. Let $P=(0, a)$ and $Q=(1, b)$, where each of $a$ and $b$ is randomly chosen from -1 to 1 . Since each circle has radius 1 , the two circles intersect if and only if $2 \geq P Q=\sqrt{(1-0)^{2}+(b-a)^{2}}$, or $-\sqrt{3} \leq b-a \leq \sqrt{3}$.

On the Cartesian plane (with axes labelled $a$ and $b$ ), the set of possible outcomes is the square bounded by the lines $a= \pm 1$ and $b= \pm 1$. The set of favourable outcomes is the region of the square between the lines $b=a-\sqrt{3}$ and $b=a+\sqrt{3}$, as shown.

Consider the non-shaded triangular part of the square in the lower right hand corner. It is easy to find that the coordinates of the vertices of the triangle are $(1,1-\sqrt{3}),(1,-1)$ and $(\sqrt{3}-1,-1)$. Thus it is a rightangled isosceles triangle with leg $2-\sqrt{3}$ and whose area is $\frac{1}{2}(2-\sqrt{3})^{2}=\frac{7-4 \sqrt{3}}{2}$. The same is true for the non-shaded triangular part of the square in the upper left hand corner. Hence the required probability is $\frac{4-(7-4 \sqrt{3})}{4}=\frac{4 \sqrt{3}-3}{4}$.