2013 AMC 10B 第 11 题

Question

实数

x

和

y

满足方程

x^2+y^2=10x-6y-34\text{。}

求

x+y

的值。

Real numbers

x

and

y

satisfy the equation

x^2+y^2=10x-6y-34.

What is

x+y?

1

2

3

6

8

知识点：配方法

把所有项移到一边并配方。

Move all terms to one side and complete two squares

两个平方数之和为零时，每个平方数都必须为零。

A sum of squares equals zero only when both squares are zero

方程可化为

x^2-10x+y^2+6y+34=0\text{。}

配方得

x^2-10x+25+y^2+6y+9

=(x-5)^2+(y+3)^2

=0\text{。}

两个平方项都非负，而它们的和为

0

，所以两者都等于

0

，即

(x-5)^2 = (y+3)^2=0\text{。}

因此

x=5,y=-3

，所以

x+y=2

。

所以正确答案是 B。

This can be rewritten as

x^2-10x+y^2+6y+34=0.

Completing the square yields

x^2-10x+25+y^2+6y+9

=(x-5)^2+(y+3)^2

=0.

Since both squared terms are nonnegative and their sum is

0,

both must be

0

. Thus

(x-5)^2 = (y+3)^2=0.

Therefore,

x=5,y=-3,

so

x+y=2.

Thus, the correct answer is B.

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