2022 AMC 10B Problem 22

Let $O$ be the origin; the first two circles have radii $2$ and $8$. A circle of radius $r$ with center at distance $d$ from $O$ tangent to both must lie inside the big circle (a circle outside it cannot reach the small one), so $d = 8 - r$. It is tangent to the small circle either externally, $d = r + 2$, or by enclosing it, $d = r - 2$.

• Externally: $8 - r = r + 2$, so $r = 3$ and $d = 5$.
• Enclosing: $8 - r = r - 2$, so $r = 5$ and $d = 3$.

Now impose tangency to the third circle, center $P = (5, 0)$, radius $\sqrt3$: the distance from our center to $P$ must be $r + \sqrt3$ or $r - \sqrt3$.

Case $r = 3$. The center lies on the circle of radius $5$ about $O$, which passes through $P$. A circle about $P$ of radius $\rho$ meets it in exactly two points whenever $0 < \rho < 10$; both $\rho = 3 + \sqrt3$ and $\rho = 3 - \sqrt3$ qualify. That is $4$ circles of area $9\pi$.

Case $r = 5$. The center lies on the circle of radius $3$ about $O$, and $OP = 5$. Circles of radii $3$ and $\rho$ with centers $5$ apart meet in two points when $|3 - \rho| < 5 < 3 + \rho$, i.e. $2 < \rho < 8$. Both $5 + \sqrt3 \approx 6.73$ and $5 - \sqrt3 \approx 3.27$ qualify: $4$ circles of area $25\pi$.

Sum of areas: $4 \cdot 9\pi + 4 \cdot 25\pi = 136\pi$.

The answer is $\boxed{\textbf{(E)}\ 136\pi}$.

Two concentric constraints determine the radius and the distance of the center from $O$, leaving the center free on a circle; the third tangency is another circle for the center to lie on, and two circles generically meet twice. Each valid $(r, \pm\sqrt3)$ combination therefore contributes two circles, mirror images across the $x$-axis by symmetry. Counting configurations rather than solving for coordinates is the whole game.

• Considering only circles squeezed inside the annulus (radius 3) and missing the radius-5 circles that enclose the small circle, giving 4 · 9π = 36π or 68π.
• Counting only external tangency to the third circle and forgetting the internally tangent option, halving each case.