Question

设 $S$ 为坐标平面中同时与三个圆 $x^2 + y^2 = 4$ 、 $x^2 + y^2 = 64$ 和 $(x - 5)^2 + y^2 = 3$ 相切的所有圆的集合。 $S$ 中所有圆的面积之和是多少？

Let $S$ be the set of circles in the coordinate plane that are tangent to each of the three circles with equations $x^2 + y^2 = 4,$ $x^2 + y^2 = 64,$ and $(x - 5)^2 + y^2 = 3.$ What is the sum of the areas of all circles in $S?$

$48\pi$

$68\pi$

$96\pi$

$102\pi$

$136\pi$

答案：E

知识点：相切圆圆面积分类讨论

难度评级：2170

解答：

前两个圆同心，半径分别为 $2$ 和 $8$ 。同时与它们相切的圆，要么半径为 $3$ 圆心到原点距离为 $5$ ；要么半径为 $5$ 圆心到原点距离为 $3$ 。

第三个圆的圆心为 $(5, 0)$ ，半径为 $\sqrt3$ 。再要求与它相切时，半径为 $3$ 的圆恰有四个，半径为 $5$ 的圆也恰有四个可行。 $s\in\{3,5\}$ ， $(5,0)$ $s+\sqrt3$ $s-\sqrt3$ 。

面积和为 $4 \cdot \pi(3)^2 + 4 \cdot \pi(5)^2$ $= 36\pi + 100\pi = 136\pi$ 。

所以正确答案是 E。

The first two circles are concentric with radii $2$ and $8.$ A circle tangent to both either has radius $3$ with center at distance $5$ from the origin, or radius $5$ with center at distance $3$ from the origin.

The third circle has center $(5, 0)$ and radius $\sqrt3.$ For each candidate radius $s\in\{3,5\},$ tangency requires the center's distance from $(5,0)$ to be $s+\sqrt3$ or $s-\sqrt3.$ Each of these two distance circles intersects the appropriate center-locus in two symmetric points. Hence exactly four radius- $3$ circles and four radius- $5$ circles work.

The sum of the areas is $4 \cdot \pi(3)^2 + 4 \cdot \pi(5)^2$ $= 36\pi + 100\pi = 136\pi.$

Thus, the correct answer is E.

2022 AMC 12B 第 21 题

答案：E

难度评级：2170

解答：

其他年份的第 21 题