Question

一个圆的圆心为 $(0, k)$ ，其中 $k \gt 6$ 。该圆与直线 $y = x$ 、 $y = -x$ 和 $y = 6$ 都相切。该圆的半径是多少？

A circle having center $(0, k),$ with $k \gt 6,$ is tangent to the lines $y = x,$ $y = -x$ and $y = 6.$ What is the radius of this circle?

$6\sqrt{2} - 6$

$6$

$6\sqrt{2}$

$12$

$6 + 6\sqrt{2}$

圆与 $y = 6$ 相切，且圆心 $(0, k)$ 在其上方，所以半径为 $r = k - 6$ 。

点 $(0, k)$ 到直线 $x - y = 0$ 的距离为 $\dfrac{|0 - k|}{\sqrt2} = \dfrac{k}{\sqrt2}$ ，这也等于 $r$ 。

令 $\dfrac{k}{\sqrt2} = k - 6$ ，得 $k = \dfrac{6\sqrt2}{\sqrt2 - 1}$ $= 6\sqrt2\,(\sqrt2+1)$ $= 12 + 6\sqrt2$ 。

因此 $r = k - 6 = 6 + 6\sqrt2$ 。

所以正确答案是 E。

Since the circle is tangent to $y = 6$ and its center $(0, k)$ is above that line, the radius is $r = k - 6.$

The distance from $(0, k)$ to the line $x - y = 0$ is $\dfrac{|0 - k|}{\sqrt2} = \dfrac{k}{\sqrt2},$ and this must also equal $r.$

Setting $\dfrac{k}{\sqrt2} = k - 6$ gives $k = \dfrac{6\sqrt2}{\sqrt2 - 1}$ $= 6\sqrt2\,(\sqrt2+1)$ $= 12 + 6\sqrt2.$

Then $r = k - 6 = 6 + 6\sqrt2.$

Thus, the correct answer is E.