图中所示为方程为 $(x+1)^2+(y-2)^2=85$ 的圆和方程为 $y=3x-20$ 的直线。直线与圆交于点 $A$ 和 $B$，圆心为 $C$。

(a) 通过计算求 $A$ 和 $B$ 的坐标。

(b) 求一个圆的方程，该圆的圆心为 $C$，且方程为 $y=3x-20$ 的直线是该圆的切线。

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。

把直线方程代入圆，解二次方程。根据原图给上、下两个交点命名。

Substitution gives $(x+1)^2+(3x-22)^2=85$, or

At $x=5$ the line has height $-5$, and at $x=8$ it has height $4$. The upper point is $A=(8,4)$ and the lower point is $B=(5,-5)$.

同心圆与直线相切时，半径等于圆心到直线的垂直距离。先用垂线斜率求切点，再用距离公式求半径。

The centre is $C=(-1,2)$. The perpendicular from $C$ to the tangent has gradient $-1/3$, so its equation is $y-2=-(x+1)/3$. At the foot $T$, it meets $y=3x-20$:

Thus $T=(13/2,-1/2)$. The squared radius is

Thus the tangent circle is $(x+1)^2+(y-2)^2=125/2$.

也可以用 (a) 的弦中点。原圆心到弦中点的线段垂直于弦，故中点就是新圆的切点。

The midpoint of $AB$ is $(13/2,-1/2)$. Its squared distance from $C$ is $(15/2)^2+(-5/2)^2=125/2$. Therefore the new circle is $(x+1)^2+(y-2)^2=125/2$.