Question

在 $xy$ -平面中，一个半径为 $4$ 、圆心在正 $x$ -轴上的圆在原点处与 $y$ -轴相切；另一个半径为 $10$ 、圆心在正 $y$ -轴上的圆在原点处与 $x$ -轴相切。经过这两个圆的两个交点的直线斜率是多少？

In the $xy$ -plane, a circle of radius $4$ with center on the positive $x$ -axis is tangent to the $y$ -axis at the origin, and a circle with radius $10$ with center on the positive $y$ -axis is tangent to the $x$ -axis at the origin. What is the slope of the line passing through the two points at which these circles intersect?

$\dfrac{2}{7}$

$\dfrac{3}{7}$

$\dfrac{2}{\sqrt{29}}$

$\dfrac{1}{\sqrt{29}}$

$\dfrac{2}{5}$

答案：E

知识点：圆根轴斜率

难度评级：1440

解答：

两个圆分别为 $(x-4)^2+y^2=16$ 和 $x^2+(y-10)^2=100$ ，即 $x^2+y^2=8x$ 和 $x^2+y^2=20y$ 。相减得 $8x=20y$ ，所以交点都在 $y=\tfrac{2}{5}x$ 上，该直线的斜率为 $\tfrac{2}{5}$ 。

因此，正确答案是 E。

The circles are $(x-4)^2+y^2=16$ and $x^2+(y-10)^2=100,$ i.e. $x^2+y^2=8x$ and $x^2+y^2=20y.$ Subtracting gives $8x=20y,$ so the intersection points lie on $y=\tfrac{2}{5}x,$ which has slope $\tfrac{2}{5}.$

Thus, the correct answer is E.

2023 AMC 12B 第 10 题

答案：E

难度评级：1440

解答：

其他年份的第 10 题