2019 AMC 12B Problem 20

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20.

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

83π8\dfrac{83\pi}{8}

21π2\dfrac{21\pi}{2}

85π8\dfrac{85\pi}{8}

43π4\dfrac{43\pi}{4}

87π8\dfrac{87\pi}{8}

Answer: C
Concepts:circletangent linecoordinate geometry
Difficulty rating: 2050
Solution:

Let P=(x,0)P=(x,0) be the intersection. Equal tangent lengths give PA=PB,PA=PB, so (x6)2+132(x-6)^2+13^2 =(x12)2+112,=(x-12)^2+11^2, yielding x=5x=5 and P=(5,0).P=(5,0).

The center O=(h,k)O=(h,k) satisfies OAPAOA\perp PA and OBPB.OB\perp PB. With PA=(1,13)PA=(1,13) and PB=(7,11),PB=(7,11), these give h+13k=175h+13k=175 and 7h+11k=205,7h+11k=205, so O=(374,514).O=\left(\dfrac{37}{4},\dfrac{51}{4}\right).

Then r2=OA2r^2=OA^2 =(134)2+(14)2=\left(\dfrac{13}{4}\right)^2+\left(\dfrac14\right)^2 =17016=858,=\dfrac{170}{16}=\dfrac{85}{8}, so the area is 85π8.\dfrac{85\pi}{8}.

Thus, C is the correct answer.

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