图中所示为方程为 $(x-2)^2+(y+4)^2=20$、圆心为 $C$ 的圆。点 $B$ 的坐标为 $(0,2)$，线段 $BC$ 与圆交于点 $P$。

(a) 求 $BC$ 的方程。

(b) 由此求 $P$ 的坐标，答案写成精确形式。

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。

由标准圆方程读出圆心。用 $B$ 和圆心求斜率与截距。

The centre is $C=(2,-4)$. The gradient from $B=(0,2)$ to $C$ is $(-4-2)/(2-0)=-3$. Hence $BC$ has equation $y=2-3x$.

把直线代入圆，得到两个交点。$P$ 位于线段 $BC$ 内，因此选横坐标介于 $0$ 和 $2$ 的交点。

Substituting $y=2-3x$ gives

Thus $x=2\pm\sqrt2$. Only $2-\sqrt2$ lies between $0$ and $2$. Its height is $2-3(2-\sqrt2)=-4+3\sqrt2$, so $P=(2-\sqrt2,-4+3\sqrt2)$.

也可以由相似三角形求坐标。$CP$ 是半径，将 $CB$ 的水平和竖直位移按相同比例缩短。

From (a), $P$ lies on the line through $B=(0,2)$ and $C=(2,-4)$. The length of $CB$ is $\sqrt{(-2)^2+6^2}=2\sqrt{10}$, while $CP$ is the radius $2\sqrt5$. Hence $CP/CB=1/\sqrt2$.

The right triangles formed by the horizontal and vertical displacements from $C$ to $P$ and from $C$ to $B$ are similar. Towards $B$, the horizontal change is $-2$ and the vertical change is $6$, so

Thus $P=(2-\sqrt2,-4+3\sqrt2)$.