AMC12 2005 B

AMC12 2005 B · Q14

AMC12 2005 B · Q14. It mainly tests Circle theorems, Coordinate geometry.

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

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

(A) 6\sqrt{2} - 6 6\sqrt{2} - 6

(B) 6 6

(D) 12 12

(E) 6 + 6\sqrt{2} 6 + 6\sqrt{2}

Answer

Correct choice: (E)

正确答案：（E）

Solution

Let $R$ be the radius of the circle. Draw the two radii that meet the points of tangency to the lines $y = \pm x$. We can see that a square is formed by the origin, two tangency points, and the center of the circle. The side lengths of this square are $R$ and the diagonal is $k = R+6$. The diagonal of a square is $\sqrt{2}$ times the side length. Therefore, $R+6 = R\sqrt{2} \Rightarrow R = \dfrac{6}{\sqrt{2}-1} = 6+6\sqrt{2} \Rightarrow \boxed{\mathrm{E}}$.

设该圆的半径为 $R$。作两条半径分别连到与直线 $y = \pm x$ 的切点。可以看出，由原点、两个切点以及圆心构成一个正方形。该正方形的边长为 $R$，对角线为 $k = R+6$。正方形的对角线长度是边长的 $\sqrt{2}$ 倍。因此，$R+6 = R\sqrt{2} \Rightarrow R = \dfrac{6}{\sqrt{2}-1} = 6+6\sqrt{2} \Rightarrow \boxed{\mathrm{E}}$。

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