2017 AMC 12B 第 9 题

Question

一个圆的圆心为

(-10, -4)

，半径为

13

。另一个圆的圆心为

(3, 9)

，半径为

\sqrt{65}

。经过这两个圆交点的直线方程为

x + y = c

。

c

是多少？

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?

3

3\sqrt{3}

4\sqrt{2}

6

\dfrac{13}{2}

将两个圆写为

(x+10)^2 + (y+4)^2 = 169

和

(x-3)^2 + (y-9)^2 = 65

Write both circles as

(x+10)^2 + (y+4)^2 = 169

and

(x-3)^2 + (y-9)^2 = 65

展开后用一个方程减去另一个方程；二次项会相消

Subtract one expanded equation from the other; the quadratic terms cancel

两个圆为

(x+10)^2 + (y+4)^2 = 169

和

(x-3)^2 + (y-9)^2 = 65

。展开并用第一个方程减去第二个方程，

x^2

与

y^2

项相消，化简为

x + y = 3

。任一交点都满足该方程，因此这就是经过两个交点的直线，且

c = 3

。

所以正确答案是 A。

The circles are

(x+10)^2 + (y+4)^2 = 169

and

(x-3)^2 + (y-9)^2 = 65.

Expanding and subtracting the second from the first cancels the

x^2

and

y^2

terms and simplifies to

x + y = 3.

Any intersection point satisfies this, so it is the line through both, and

c = 3.

Thus, the correct answer is A.

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