AMC12 2017 B

AMC12 2017 B · Q9

AMC12 2017 B · Q9. It mainly tests Circle theorems, Coordinate geometry.

A circle has center (−10, −4) and radius 13. Another circle has center (3, 9) and radius $\sqrt{65}$. The line passing through the two points of intersection of the two circles has equation $x + y = c$. What is $c$?

一个圆的圆心为 (−10, −4)，半径 13。另一个圆的圆心为 (3, 9)，半径 $\sqrt{65}$。通过两个圆交点连线方程为 $x + y = c$。$c$ 是多少？

(A) 3 3

(B) $3\sqrt{3}$ $3\sqrt{3}$

(D) 6 6

(E) 13 13

Answer

Correct choice: (A)

正确答案：（A）

Solution

Answer (A): The first circle has equation $(x+10)^2+(y+4)^2=169$, and the second circle has equation $(x-3)^2+(y-9)^2=65$. Expanding these two equations, subtracting, and simplifying yields $x+y=3$. Because the points of intersection of the two circles must satisfy this new equation, it must be the required equation of the line through those points, so $c=3$. In fact, the circles intersect at $(2,1)$ and $(-5,8)$.

答案（A）：第一个圆的方程为 $(x+10)^2+(y+4)^2=169$，第二个圆的方程为 $(x-3)^2+(y-9)^2=65$。展开这两个方程，相减并化简得到 $x+y=3$。因为两圆的交点必须满足这个新方程，所以它就是经过这些交点的直线方程，因此 $c=3$。事实上，两圆交于 $(2,1)$ 和 $(-5,8)$。

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