AMC12 2019 B

AMC12 2019 B · Q20

AMC12 2019 B · Q20. It mainly tests Circle theorems, Coordinate geometry.

Points A(6,13) and B(12,11) lie on a circle $\omega$ in the plane. Suppose that the tangent lines to $\omega$ at A and B intersect at a point on the x-axis. What is the area of $\omega$?

点A(6,13)和B(12,11)在平面上的圆$\omega$上。假设$\omega$在A和B处的切线相交于x轴上某点。$\omega$的面积是多少？

(A) $\frac{83\pi}{8}$ $\frac{83\pi}{8}$

(B) $\frac{21\pi}{2}$ $\frac{21\pi}{2}$

(D) $\frac{43\pi}{4}$ $\frac{43\pi}{4}$

(E) $\frac{87\pi}{8}$ $\frac{87\pi}{8}$

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Let $T$ be the point where the tangents at $A$ and $B$ intersect. By symmetry $T$ lies on the perpendicular bisector $\ell$ of $AB$, so in fact $T$ is the (unique) intersection point of line $\ell$ with the $x$-axis. Computing the midpoint $M$ of $AB$ gives $(9,12)$, and computing the slope of $AB$ gives $\frac{13-11}{6-12}=-\frac{1}{3}$. This means that the slope of $\ell$ is $3$, so the equation of $\ell$ is given by $y-12=3(x-9)$. Setting $y=0$ yields that $T=(5,0)$.

答案（C）：设 $T$ 为过点 $A$ 与 $B$ 的切线的交点。由对称性，$T$ 位于线段 $AB$ 的垂直平分线 $\ell$ 上，因此 $T$ 实际上是直线 $\ell$ 与 $x$ 轴的（唯一）交点。计算 $AB$ 的中点 $M$ 得到 $(9,12)$，计算 $AB$ 的斜率得到 $\frac{13-11}{6-12}=-\frac{1}{3}$。这意味着 $\ell$ 的斜率为 $3$，因此 $\ell$ 的方程为 $y-12=3(x-9)$。令 $y=0$ 可得 $T=(5,0)$。

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