图中显示方程为 $x^2+y^2=20$ 的圆。在 $B$ 点和 $C$ 点处与圆相切的两条切线均经过点 $A(0,10)$。

(a) 设一条切线的方程为 $y=mx+10$，求 $m$ 的两个可能值。

(b) 求 $B$ 和 $C$ 的坐标。

点 $D$ 是圆与 $x$ 轴正半轴的交点。

(c) 求角 $BDC$ 的度数。

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。

按题意代入 $y=mx+10$。相切时关于 $x$ 的交点方程有重根。

Substituting in the circle gives

Its discriminant must vanish, so

Thus the possible gradients are $m=2$ and $m=-2$.

用重根的公式求出切点的横坐标，再代入切线。按原图给左右切点命名。

The double root is $x=-10m/(1+m^2)=-2m$ for $m=\pm2$. The corresponding height is $y=m(-2m)+10=2$. The left point is $B=(-4,2)$ and the right point is $C=(4,2)$.

半径与切线垂直。利用过圆心的法线方程与切线联立，求出两个切点。

For the tangent $y=2x+10$, the normal through the origin is $y=-x/2$. At their intersection,

For the tangent $y=-2x+10$, the normal is $y=x/2$. Hence

Thus $B=(-4,2)$ and $C=(4,2)$.

用圆心处的对称关系求弧 $BC$ 的圆心角，再取圆周角的一半。

Since $OB$ goes four units left and two up from the centre, $\angle BOA=\arctan(4/2)$. The points $B$ and $C$ are symmetric about $OA$, so the minor central angle $BOC$ is $2\arctan2$. The point $D$ lies outside the minor arc $BC$. Therefore $\angle BDC=\tfrac12\angle BOC=63.4^\circ$ to one decimal place.

也可以用三角形 $BDC$ 的三条边直接求角。先由圆与正横轴的交点确定 $D$。

We have $D=(\sqrt{20},0)$ and $BC=8$. The squared lengths are

Their product gives $BD\cdot CD=8\sqrt5$. By the cosine rule,

Hence $\angle BDC=63.4^\circ$ to one decimal place.

从D引水平向左和竖直向上的辅助方向。分别求DB与左向水平、DC与竖直的夹角，再从直角中扣除。

Since $D=(2\sqrt5,0)$, let $\beta$ be the angle of $DB$ above the horizontal ray pointing left, and let $\gamma$ be the angle of $DC$ to the left of the vertical ray pointing up. The coordinate differences give

Both ratios equal $\sqrt5-2$, so $\beta=\gamma=13.282525\ldots^\circ$. The required angle is

to one decimal place.