Question

两个圆中，较大圆的圆心为 $P$ ，半径为 $p$ ；较小圆的圆心为 $Q$ ，半径为 $q$ 。连接 $\overline{PQ}$ 。下列哪个说法是错误的？

Given the larger of two circles with center $P$ and radius $p$ and the smaller with center $Q$ and radius $q.$ Draw $\overline{PQ}.$ Which of the following statements is false?

$p-q$ 可以等于 $\overline{PQ}$

$p-q$ can be equal to $\overline{PQ}$

$p+q$ 可以等于 $\overline{PQ}$

$p+q$ can be equal to $\overline{PQ}$

$p+q$ 可以小于 $\overline{PQ}$

$p+q$ can be less than $\overline{PQ}$

$p-q$ 可以小于 $\overline{PQ}$

$p-q$ can be less than $\overline{PQ}$

以上说法均非错误

none of these

答案：E

知识点：circles center distance tangency

难度评级：1400

小提示：

分别考虑两圆内切、外切和相离时的圆心距

Interpret the center distance for internal tangency, external tangency, and disjoint circles

大提示：

对前四个说法逐一尝试构造一种可行的两圆位置关系

For each of the first four statements, try to choose a valid relative position of the circles

解答：

两圆内切时有 $PQ=p-q$ ，外切时有 $PQ=p+q$ 。两圆相离时可以有 $PQ\gt p+q$ ，而许多相交或相离的情形满足 $PQ\gt p-q$ 。因此 A 至 D 中的每种情况都可能发生，没有一个说法是错误的。

因此，正确答案是 E。

Internal tangency realizes $PQ=p-q,$ and external tangency realizes $PQ=p+q.$ Disjoint circles can have $PQ\gt p+q,$ and many intersecting or disjoint configurations have $PQ\gt p-q.$ Thus each of A through D can occur, so none of them is false.

Therefore, the correct answer is E.

1953 AMC 12 第 14 题

答案：E

难度评级：1400

小提示：

大提示：

解答：

其他年份的第 14 题