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 $m \gt 0$ 。求 $b$ ？

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

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 $\left(-\tfrac{38}{5}, 0\right)$ 。

另一条切线的倾角为 $2\theta$ ，这里以 $x$ -轴为基准，所以其斜率为因此 $b = \tfrac{120}{119} \cdot \tfrac{38}{5} = \tfrac{912}{119}$ 。 $\tan 2\theta = \frac{2 \cdot \tfrac{5}{12}}{1 - \left(\tfrac{5}{12}\right)^2} = \frac{120}{119}.$

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

解答：

其他年份的第 19 题