2022 AMC 10B Problem 22

The first two circles are centred at the origin $O$ with radii $2$ and $8$. The third is centred at $P = (5, 0)$ with radius $\sqrt3$. Start with the first two, because they fix the size of the circle we want.

1. Where the circle must be. A circle touching both the radius-$8$ circle and the radius-$2$ circle must sit inside the big one, touching it from the inside. (A circle outside the big one could never reach the small one.) So if it has radius $r$ and centre $C$, then $OC = 8 - r$.
2. Touching the small circle. Either it sits beside the small circle (touching from outside), so $OC = r + 2$, or it wraps around the small circle (touching from inside), so $OC = r - 2$.
3. Beside: $8 - r = r + 2$ gives $r = 3$, and $OC = 5$.
4. Wrapped around: $8 - r = r - 2$ gives $r = 5$, and $OC = 3$.
5. Now the third circle. Our circle touches the circle around $P$ either from outside, so $CP = r + \sqrt3$, or by wrapping around it, so $CP = r - \sqrt3$. (Our $r$ is $3$ or $5$, both bigger than $\sqrt3 \approx 1.73$, so the small circle cannot wrap around ours.)
6. Radius 3. The centre $C$ lies on the circle of radius $5$ around $O$, which passes through $P$. We need $CP = 3 + \sqrt3 \approx 4.73$ or $CP = 3 - \sqrt3 \approx 1.27$. On a circle of radius $5$, points at distance $\rho$ from the point $P$ on it exist in pairs (one above the $x$-axis, one below) for every $\rho$ between $0$ and $10$. So each distance gives $2$ centres: $4$ circles of area $9\pi$.
7. Radius 5. The centre $C$ lies on the circle of radius $3$ around $O$, and $P$ is $5$ from $O$. The nearest point of that circle to $P$ is $5 - 3 = 2$ away and the farthest is $5 + 3 = 8$ away. We need $CP = 5 + \sqrt3 \approx 6.73$ or $5 - \sqrt3 \approx 3.27$, both strictly between $2$ and $8$, so again each gives $2$ centres: $4$ circles of area $25\pi$.
8. Add the areas. $4 \times 9\pi + 4 \times 25\pi = 36\pi + 100\pi = 136\pi$.

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

Two circles with the same centre decide both the size of a touching circle and how far its centre is from that common centre, so the centre must lie on a known circle. Touching the third circle puts the centre at a fixed distance from another point, which is a second circle, and two circles usually cross at two points. Remember both ways of touching: side by side, and one around the other.

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