1996 AMC 12 第 18 题

Question

一个半径为

2

的圆，圆心为

(2,0)

。另一个半径为

1

的圆，圆心为

(5,0)

。一条直线在第一象限内的两点分别与这两个圆相切。下列哪一个数最接近该直线的

y

轴截距？

A circle of radius

2

has center at

(2,0).

A circle of radius

1

has center at

(5,0).

A line is tangent to the two circles at points in the first quadrant. Which of the following is closest to the

y

-intercept of the line?

\frac{\sqrt2}{4}

\frac{\sqrt8}{3}

1+\sqrt3

2\sqrt2

3

把切线写成

y=mx+b

，并对两个圆心分别使用点到直线的距离公式

Write the tangent as

y=mx+b

and use point-to-line distance for each center

先将两个距离方程相减求出斜率，再求截距

Subtract the two distance equations to determine the slope before solving for the intercept

把上方的公切线写成

mx-y+b=0

，并令

s=\sqrt{m^2+1}

。两个圆心到直线的距离给出

2m+b=2s,\qquad 5m+b=s\text{。}

因此

3m=-s

，所以

9m^2=m^2+1

，且

m=-\frac{1}{\sqrt8}

。由第二个方程，

\begin{aligned} b&=s-5m\\ &=\frac3{\sqrt8}+\frac5{\sqrt8}=2\sqrt2 \end{aligned}\text{。}

所以正确答案是 D。

Write the upper common tangent as

mx-y+b=0

and let

s=\sqrt{m^2+1}.

The two center-to-line distances give

2m+b=2s,\qquad 5m+b=s.

Thus

3m=-s,

so

9m^2=m^2+1

and

m=-\frac{1}{\sqrt8}.

From the second equation,

\begin{aligned} b&=s-5m\\ &=\frac3{\sqrt8}+\frac5{\sqrt8}=2\sqrt2. \end{aligned}

Therefore the correct answer is D.

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