2022 AMC 10B Problem 22

P21-25~7 minPrint

Circles Coordinate Geometry Split into exhaustive cases and handle each Exploit symmetry to reduce work or pair up objects

Let $S$ be the set of circles in the coordinate plane that are tangent to each of the three circles with equations $x^{2}+y^{2}=4$ , $x^{2}+y^{2}=64$ , and $(x-5)^{2}+y^{2}=3$ . What is the sum of the areas of all circles in $S$ ?

Show solution without answering

Key insight

Touching both circles centred at the origin forces radius 3 or radius 5. Each size can then touch the third circle in four positions.

Solution

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.

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 $OC = 8 - r$ .
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 $OC = r - 2$ .
Beside: $8 - r = r + 2$ gives $r = 3$ , and $OC = 5$ .
Wrapped around: $8 - r = r - 2$ gives $r = 5$ , and $OC = 3$ .
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.)
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 $9\pi$ .
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 $25\pi$ .
Add the areas. $4 \times 9\pi + 4 \times 25\pi = 36\pi + 100\pi = 136\pi$ .

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

Why this works

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.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Geometry, Chapter 11: Circles
Introduction to Geometry, Chapter 17: Analytic Geometry
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.2)
Introduction to Algebra, Chapter 22: Special Manipulations (§22.3)