1978 AMC 12 第 11 题

Question

若

r

为正数，且直线

x+y=r

与圆

x^2+y^2=r

相切，则

r

等于

If

r

is positive and the line whose equation is

x+y=r

is tangent to the circle whose equation is

x^2+y^2=r,

then

r

equals

\frac12

1

2

\sqrt2

2\sqrt2

该圆的圆心在原点，半径为

\sqrt r

The circle has center at the origin and radius

\sqrt r

令原点到直线的距离等于圆的半径

Set the distance from the origin to the line equal to the circle’s radius

原点到直线

x+y-r=0

的距离为

\frac{r}{\sqrt2}

。相切要求该距离等于半径

\sqrt r

，所以

\frac r{\sqrt2}=\sqrt r\text{。}

由于

r\gt0

，两边平方再除以

r

，得到

r=2

。

因此，正确答案是 C。

The distance from the origin to

x+y-r=0

is

\frac{r}{\sqrt2}.

Tangency requires this to equal the radius

\sqrt r,

so

\frac r{\sqrt2}=\sqrt r.

Since

r\gt0,

squaring and dividing by

r

gives

r=2.

Therefore, the correct answer is C.

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