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2004 AMC 10B Problem 12

2004 AMC 10B Problem 12AMC 10 Step by Step

Name: Date: September 13, 2026

Circles · difficulty 2 of 5 · about 2 min · position P11-15

An annulus is the region between two concentric circles. The concentric circles in the figure have radii $b$ and $c$ , with $b>c$ . Let $OX$ be a radius of the larger circle, let $XZ$ be tangent to the smaller circle at $Z$ , and let $OY$ be the radius of the larger circle that contains $Z$ . Let $a=XZ$ , $d=YZ$ , and $e=XY$ . What is the area of the annulus?

A)
$\pi a^2$
B)
$\pi b^2$
C)
$\pi c^2$
D)
$\pi d^2$
E)
$\pi e^2$

Answer

Key insight

A radius meets a tangent at a right angle, so a^2 + c^2 = b^2 and the annulus area pi(b^2 - c^2) equals pi a^2.

Solution

The annulus is the big disk minus the small disk, so its area is
$\pi b^2 - \pi c^2 = \pi\,(b^2 - c^2).$

Now relate $b^2 - c^2$ to the labeled segments. $XZ$ is tangent to the small circle at $Z$ , and $OZ$ is a radius of that circle, so $OZ \perp XZ$ . Triangle $OZX$ is therefore right-angled at $Z$ with legs $OZ = c$ and $ZX = a$ and hypotenuse $OX = b$ . The Pythagorean theorem gives
$a^2 + c^2 = b^2 \quad\Longrightarrow\quad b^2 - c^2 = a^2.$

Hence the annulus has area $\pi a^2$ .

The answer is $\boxed{\textbf{(A)}\ \pi a^2}$ .

Why this works

The "tangent is perpendicular to the radius" fact is the standard way to manufacture a right triangle inside a circle problem, and $b^2 - c^2$ is exactly the difference of squares that a right triangle produces. The distractors $d$ and $e$ are real lengths in the figure but have no simple relation to the area; the tangent segment is the one tied to both radii at once.

Alternative approach

Sanity check with extremes: if $c$ grows toward $b$ , the annulus shrinks to nothing, and so do $a$ , $d$ , and $e$ , ruling out (B) and (C). Sending $c$ to $0$ instead turns the annulus into the whole disk and makes $Z = O$ , so $a = b$ and (A) gives the right area, while $d = e = b$ would make (D) and (E) plausible too; only the Pythagorean relation settles it in general.

The trap

Missing the right angle at Z (radius perpendicular to tangent) and trying to build the answer from the segments d or e instead.

Common mistakes

Missing the right angle at Z (radius perpendicular to tangent) and trying to build the answer from the segments d or e instead.
Writing the Pythagorean relation as $b^2 + c^2 = a^2$ , treating $XZ$ as the hypotenuse.

Techniques

Use the answer choices (mod checks, size, form) to eliminate or select · Set up the equation/formula and compute; no special trick needed