AMC12 2023 B

AMC12 2023 B · Q10

AMC12 2023 B · Q10. It mainly tests Circle theorems, Coordinate geometry.

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?

在$xy$平面中，一个半径为4、圆心在正$x$轴上的圆与$y$轴在原点相切，一个半径为10、圆心在正$y$轴上的圆与$x$轴在原点相切。这两个圆相交的两点的连线斜率是多少？

(A) \ \dfrac{2}{7} \ \dfrac{2}{7}

(B) \ \dfrac{3}{7} \ \dfrac{3}{7}

(D) \ \dfrac{1}{\sqrt{29}} \ \dfrac{1}{\sqrt{29}}

(E) \ \dfrac{2}{5} \ \dfrac{2}{5}

Answer

Correct choice: (E)

正确答案：（E）

Solution

The center of the first circle is $(4,0)$. The center of the second circle is $(0,10)$. Thus, the slope of the line that passes through these two centers is $- \frac{10}{4} = - \frac{5}{2}$. Because this line is the perpendicular bisector of the line that passes through two intersecting points of two circles, the slope of the latter line is $\frac{-1}{- \frac{5}{2}} = \boxed{\textbf{(E) } \frac{2}{5}}$.

第一个圆的圆心为$(4,0)$。第二个圆的圆心为$(0,10)$。因此，通过这两个圆心的直线的斜率为$- \frac{10}{4} = - \frac{5}{2}$。因为这条线是两个圆相交两点连线的垂直平分线，后者的斜率为$\frac{-1}{- \frac{5}{2}} = \boxed{\textbf{(E) } \frac{2}{5}}$。

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