图中所示的是一个圆及其两条切线 $L_1$ 和 $L_2$。切线 $L_1$ 与 $x$ 轴交于 $Q$，与 $y$ 轴交于 $P$；切线 $L_2$ 与 $x$ 轴交于 $S$，与 $y$ 轴交于 $R$。

(a) 圆的一条直径的两个端点为 $(0,4-\sqrt{10})$ 和 $(0,4+\sqrt{10})$。

证明该圆的方程可写为

(b) 两条切线 $L_1$ 和 $L_2$ 的方程均具有 $y=-3x+c$ 的形式。

证明 $c^2-8c-84=0$。

(c) 求四边形 $PQRS$ 面积的精确值。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

直径端点的中点是圆心，半径是直径的一半。

The given endpoints have midpoint $(0,4)$ and are separated by $2\sqrt{10}$. Thus the radius is $\sqrt{10}$ and the equation is $x^2+(y-4)^2=10$. Expanding and rearranging gives

把切线的表达式代入圆。相切时，关于 $x$ 的二次方程有一个重根。

Substitution of $y=-3x+c$ gives

Since the line is tangent, the discriminant vanishes:

Dividing by $-4$ yields $c^2-8c-84=0$.

也可以用圆心到切线的距离等于半径来建立方程。

The line is $3x+y-c=0$. Its distance from $(0,4)$ is $|4-c|/\sqrt{10}$. Equating this to the radius $\sqrt{10}$ gives $|4-c|=10$. Thus $(c-4)^2=100$, which expands to $c^2-8c-84=0$.

过切点的半径斜率为 $1/3$。先把这条直径与圆联立，得到两个切点，再求各自的切线截距。

The circle has centre $(0,4)$ and radius $\sqrt{10}$. A tangent with slope $-3$ has a perpendicular radius with slope $1/3$. The diameter through the points of contact is therefore $y=x/3+4$. Substituting into $x^2+(y-4)^2=10$ gives

The points of contact are $(3,5)$ and $(-3,3)$. In $y=-3x+c$, they give $c=5+9=14$ and $c=3-9=-6$, respectively. Thus the two possible values of $c$ satisfy

将上下半圆分别写成显函数，用链式法则找斜率为 $-3$ 的点。平方前先看横坐标的符号，避免把两支的切点混用。

For the upper branch, $y=4+\sqrt{10-x^2}$ and

Setting the gradient equal to $-3$ gives $x=3\sqrt{10-x^2}>0$. Squaring gives $x^2=9(10-x^2)$, hence $x=3$ and $y=5$. This gives $c=5+3(3)=14$.

For the lower branch, $y=4-\sqrt{10-x^2}$ and

The gradient equation is $x=-3\sqrt{10-x^2}<0$. It again gives $x^2=9$, but now $x=-3$ and $y=3$. Thus $c=3+3(-3)=-6$. Both points satisfy their unsquared gradient equations. Therefore

用 (b) 求两条切线的截距。以 $PR$ 为公共底，将四边形分成左右两个三角形。

Factoring gives $(c-14)(c+6)=0$. From the diagram, the positive intercept belongs to $L_1$, so

The base $PR$ has length $20$. The perpendicular distances of $Q$ and $S$ from the base line are $14/3$ and $2$, respectively. Therefore the exact area is

$PQ$ 与 $RS$ 分别在两条平行切线上，可作为梯形的两底。两条切线位于圆的两侧，它们之间的垂直距离就是直径。

Solving $(c-14)(c+6)=0$ gives

The parallel sides have lengths

and

The perpendicular distance between these tangents is the diameter $2\sqrt{10}$. Thus the trapezium area is