1995 AMC 12 Problem 26

Attempt Problem 26 of the 1995 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1995 AMC 12 solutions, or check the answer key.

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26.

In the figure,

\overline{AB}

and

\overline{CD}

are diameters of the circle with center

O,

\overline{AB}\perp\overline{CD},

and chord

\overline{DF}

intersects

\overline{AB}

at

E.

If

DE=6

and

EF=2,

then the area of the circle is

23\pi

\frac{47}2\pi

24\pi

\frac{49}2\pi

25\pi

Answer: C

Concepts:intersecting chords coordinates circles

Difficulty rating: 2170

Small hint:

Intersecting chords gives

AE\cdot EB=DE\cdot EF

Big hint:

Place

O

at the origin and express

F

using the fact that

E

divides

\overline{DF}

in a

3:1

ratio

Solution:

Let the radius be

r

and

OE=x.

Intersecting chords gives

(r+x)(r-x)=6\cdot2,

so

r^2-x^2=12.

Put

O=(0,0),

D=(0,-r),

and

E=(x,0).

Since

DE:EF=3:1,

\begin{aligned} F&=D+\frac43(E-D)\\ &=\left(\frac{4x}{3},\frac r3\right). \end{aligned}

Because

F

lies on the circle,

\frac{16x^2}{9}+\frac{r^2}{9}=r^2,

so

r^2=2x^2.

Therefore

x^2=12

and

r^2=24.

The area is

24\pi,

so the correct answer is C.