1998 AMC 12 第 23 题

Question

图像

x^2+y^2=4+12x+6y

与

x^2+y^2=k+4x+12y

在

k

满足

a\le k\le b

时相交，并且对其他

k

值均不相交。求

b-a

。

The graphs of

x^2+y^2=4+12x+6y

and

x^2+y^2=k+4x+12y

intersect when

k

satisfies

a\le k\le b,

and for no other values of

k.

Find

b-a.

5

68

104

140

144

配方求出两个圆的圆心和半径

Complete the square to find both circle centers and radii

两圆相交，当且仅当圆心距介于两半径之差与两半径之和之间

Two circles intersect exactly when the center distance lies between the difference and sum of their radii

两个圆分别为

(x-6)^2+(y-3)^2=49\text{，}

(x-2)^2+(y-6)^2=k+40\text{。}

两圆圆心距为

5

。设第二个圆的半径为

r=\sqrt{k+40}

，相交条件为

|7-r|\le5\le7+r

，即

2\le r\le12

。所以

-36\le k\le104

，且

b-a=140

。正确答案是 D。

The circles are

(x-6)^2+(y-3)^2=49,

(x-2)^2+(y-6)^2=k+40.

Their centers are

5

units apart. With second radius

r=\sqrt{k+40},

intersection requires

|7-r|\le5\le7+r,

or

2\le r\le12.

Thus

-36\le k\le104,

and

b-a=140.

The correct answer is D.

其他年份的第 23 题