2006 AMC 12A 第 19 题

Question

圆心为

(2, 4)

和

(14, 9)

的两个圆半径分别为

4

和

9

，两圆的一条公共外切线方程可写为

y = mx + b

，其中

m \gt 0

。

b

是多少？

Circles with centers

(2, 4)

and

(14, 9)

have radii

4

and

9,

respectively. The equation of a common external tangent to the circles can be written in the form

y = mx + b

with

m \gt 0.

What is

b?

\dfrac{908}{119}

\dfrac{909}{119}

\dfrac{130}{17}

\dfrac{911}{119}

\dfrac{912}{119}

答案：E

知识点：切线三角恒等式坐标几何

难度评级：1960

小提示：

每个圆都与

x

-轴相切，所以

x

-轴是一条公共外切线。

Each circle is tangent to the

x

-axis, so the

x

-axis is one common external tangent

大提示：

两条外切线交于圆心连线所在直线，其斜率为

\tfrac{5}{12} = \tan\theta

；另一条切线斜率为

\tan 2\theta

。

The two external tangents meet on the line through the centers, whose slope is

\tfrac{5}{12} = \tan\theta;

the other tangent has slope

\tan 2\theta

解答：

每个圆的半径等于其圆心的

y

-坐标，所以两圆都与

x

-轴相切，这条直线就是一条公共外切线。两条外切线交于圆心连线与

x

-轴的交点。

该直线斜率为

\tfrac{9 - 4}{14 - 2} = \tfrac{5}{12} = \tan\theta

，且过

(2, 4)

，与

x

-轴交于

\left(-\tfrac{38}{5}, 0\right)

。

另一条切线与

x

-轴所成的角为

2\theta

，所以其斜率为

\tan 2\theta = \frac{2 \cdot \tfrac{5}{12}}{1 - \left(\tfrac{5}{12}\right)^2} = \frac{120}{119}\text{。}

因此

b = \tfrac{120}{119} \cdot \tfrac{38}{5} = \tfrac{912}{119}

。

因此，正确答案是 E。

Each circle’s radius equals its center’s

y

-coordinate, so both are tangent to the

x

-axis, which is a common external tangent. The two external tangents meet at the

x

-intercept of the line through the centers.

That line has slope

\tfrac{9 - 4}{14 - 2} = \tfrac{5}{12} = \tan\theta

and passes through

(2, 4),

meeting the

x

-axis at

\left(-\tfrac{38}{5}, 0\right).

The other tangent makes angle

2\theta

with the

x

-axis, so its slope is

\tan 2\theta = \frac{2 \cdot \tfrac{5}{12}}{1 - \left(\tfrac{5}{12}\right)^2} = \frac{120}{119}.

Then

b = \tfrac{120}{119} \cdot \tfrac{38}{5} = \tfrac{912}{119}.

Thus, the correct answer is E.

2006 AMC 12A 第 19 题

答案：E

难度评级：1960

小提示：

大提示：

解答：

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