Three-point disk coverage with the smallest radius

Three-point disk coverage asks how small one filled disk can be while still containing three specified points. For the invented coordinates P = (−1, 0), Q = (1, 0), and R = (0, 2), the answer is radius 5/4 with center (0, 3/4). The useful part is the certificate behind that answer: one calculation checks that the disk contains everything, and a second calculation rules out every smaller radius across the entire plane.

State what the disk must contain

A closed filled disk includes its interior and its circular boundary. A point on the boundary therefore counts as covered. Write an arbitrary center as (x, y) and its radius as r, with r nonnegative. Its three coverage conditions are (x + 1)² + y² ≤ r², (x − 1)² + y² ≤ r², and x² + (y − 2)² ≤ r². All three inequalities must hold for that same center and radius.

These coordinates are stipulated mathematical inputs. They are not measurements of a room, a broom head, a cleaning reach, or an actual product. The plane has no walls, obstacles, inaccessible centers, or movement restrictions. For a chosen center, the required radius is the largest of its three point distances. The task is to minimize that largest distance while allowing both center coordinates to vary freely.

Find a candidate without mistaking it for a proof

Start by looking for a center equally distant from P and Q. Equating their squared distances gives (x + 1)² + y² = (x − 1)² + y². Subtracting the common terms leaves 4x = 0, so such a center has x = 0. This identifies the vertical line on which the first two distances agree. It does not yet establish where the best covering center lies.

On that line, equate the squared distance to P with the squared distance to R: 1 + y² = (y − 2)². Expanding the right side gives y² − 4y + 4. Cancellation yields 4y = 3, hence y = 3/4. The resulting candidate is C = (0, 3/4). Equal-distance equations have produced a sensible point to test, rather than a license to declare it optimal.

Now check every constraint directly. From C to P, the squared distance is 1 + 9/16 = 25/16. The calculation for Q is identical. From C to R, the vertical difference is 2 − 3/4 = 5/4, whose square is also 25/16. Thus radius 5/4 covers all three points, each exactly on its boundary. We have established an attainable upper bound on the smallest possible radius.

Compare two tempting centers

The unweighted average of the three coordinates is G = (0, 2/3). At G, each lower point has squared distance 1 + 4/9 = 13/9. The upper point has distance 2 − 2/3 = 4/3, whose square is 16/9. The upper point therefore controls the required covering radius: it is 4/3. Since 4/3 exceeds 5/4 by 1/12, this average-coordinate choice loses to the candidate already checked.

The midpoint of the lower pair is O = (0, 0). Its distances to P and Q are both 1, but its distance to R is 2. A disk centered there needs radius 2. Inspecting only the balanced lower pair misses the active upper constraint. Neither this failure nor the centroid comparison proves global optimality of C; they merely eliminate two particular guesses with explicit calculations.

Use a weighted mean to bound the largest distance

For an arbitrary center, name the three squared distances a, b, and c in the order P, Q, R. Give them weights 5/16, 5/16, and 3/8. These weights are positive and sum to 1. Their weighted mean W = (5/16)a + (5/16)b + (3/8)c cannot exceed the largest of a, b, and c. Each term being averaged is at most that largest value, and multiplying by positive weights preserves those inequalities.

This comparison connects an easy expression with the actual coverage objective. A covering disk must have r² at least as large as each squared distance, hence at least as large as their maximum and their weighted mean. The mean is a lower-bound certificate for the maximum; replacing the task with a mean-distance minimization would ask a different question. The distinction matters even though the certificate itself is built from an average.

Expand the certificate exactly

The first two weighted terms together are (5/16)[(x + 1)² + y² + (x − 1)² + y²]. Their opposite linear x terms cancel, leaving (5/8)x² + (5/8)y² + 5/8. The third term is (3/8)[x² + (y − 2)²], which expands to (3/8)x² + (3/8)y² − (3/2)y + 3/2. Keeping the constants as fractions makes the resulting identity easy to inspect.

Adding those expressions gives W = x² + y² − (3/2)y + 17/8. Complete the square in y: y² − (3/2)y = (y − 3/4)² − 9/16. Since 17/8 − 9/16 = 25/16, the exact identity is W = x² + (y − 3/4)² + 25/16. This identity holds for every real x and y, without assuming that any two of the three distances agree.

Both square terms are nonnegative. Consequently W is always at least 25/16, and the maximum squared distance is always at least W. Every covering disk therefore satisfies r² ≥ 25/16. Because r is nonnegative, its radius must be at least 5/4. The candidate disk attains that bound, so the upper bound from its coverage check and this global lower bound meet exactly.

Prove that the best center is unique

Suppose another center also admitted a covering radius of 5/4. Its squared radius would be 25/16, so the certificate forces x² + (y − 3/4)² + 25/16 ≤ 25/16. Subtract the common constant. A sum of two nonnegative squares can be at most zero only when both squares vanish. Thus x = 0 and y = 3/4. There is exactly one center at the minimum radius.

The proof does not infer uniqueness from a symmetric sketch or from the fact that three points touch one circle. It uses the equality condition of an inequality that applies to every possible center. The radius is unique as a minimum numerical value; the additional equality argument identifies the center uniquely as well. These are related conclusions, but they require their own explicit reasoning.

Check a pair without forgetting the third point

For P and R alone, their midpoint is M = (−1/2, 1). Each is at squared distance 1/4 + 1 = 5/4 from M. However, Q is at squared distance (3/2)² + 1 = 13/4. The smaller disk suggested by that pair leaves Q outside. A two-point calculation can be internally correct and still fail the three-point task when its omitted constraint is restored.

For the lower pair P and Q, radius 1 around O covers both, yet R again lies outside. These pair checks explain why an answer needs a complete feasibility test. The global certificate then supplies what testing pairs cannot: a lower bound that holds even for centers never proposed or drawn. No finite list of attractive guesses would provide that all-plane statement by itself.

Separate minimum-radius uniqueness from ordinary feasibility

At the larger radius 3/2, both centers U = (0, 1) and V = (0, 1/2) work. At U the lower squared distances are 2 and the upper squared distance is 1; all are at most 9/4. At V the lower squared distances are 5/4 and the upper squared distance is 9/4. Both disks therefore cover the same three points, despite having different centers.

There is no conflict with the uniqueness result. That result applies at the smallest radius, 5/4. Increasing the allowed radius changes the feasible set of centers. Likewise, fixing a center to a line before solving would introduce a restriction absent from the stated task. Here the lower certificate already ranges over the whole plane, so it needs no unproved assumption that an optimal center sits on the symmetry line.

See how displacement raises the required bound

The certificate also quantifies what happens when a proposed center moves away from C. Write its displacement as (u, v), so x = u and y = 3/4 + v. Substituting into the identity gives W = 25/16 + u² + v². Therefore any disk at that displaced center needs squared radius at least 25/16 + u² + v². This is a bound on the actual largest distance, not necessarily its exact value at that center.

For example, moving horizontally by 1/4 gives the lower bound 26/16. Direct inspection shows why the inequality can be strict: at center (1/4, 3/4), the squared distance to P is 25/16 + 9/16 = 34/16. The distances to Q and R are 18/16 and 26/16 respectively. Their weighted mean is 26/16, while their maximum is 34/16. The mean suffices to reject that center at the minimum radius, even though it understates the radius actually needed there.

This separates the two roles of the calculations. Use the three explicit distance inequalities to check feasibility at a particular center. Use the weighted identity to rule out a radius or prove uniqueness across infinitely many centers. A certificate need not reproduce the objective everywhere to give an exact optimum where its lower bound is attained.

Keep the product context within its evidence

The catalog account for the Genuine Joe GJO80161 whisk broom supplies a handheld-broom context. Its sole gallery image shows a fiber fan with three red stitch lines, a wrapped neck, and a hanging loop. Those observations do not authenticate a maker label, fiber material, dimensions, delivered quantity, or cleaning performance.

The 18 in. Indoor & Outdoor Broom Push account provides a contrasting push-broom context. Its sole image shows a long shaft, a wide gray head with green bristles, and a raised angled connector. The catalog title is retained as a title; the image does not verify its stated size or establish any pivot function or measured sweep coverage.

Neither account supplies the three invented coordinates or a physical circular reach. A manually moved broom does not acquire a disk-shaped performance region because a diagram contains a disk. The mathematical example stays useful precisely because its inputs, permitted centers, and coverage rule are declared explicitly rather than attributed to those product photographs.

Distinguish this task from nearby geometry

The guide to a closest scalar multiple studies a target relative to one allowed line and an ordinary squared-distance residual. Here the center ranges over the whole plane and must satisfy three containment constraints together. The weighted expression is only a certificate for the largest distance, preserving that different objective.

The guide to drawing a circle with a fixed center and radius begins with those two inputs already chosen. This example chooses them by optimization. A drawn circumference also differs from the filled disk used in the coverage inequalities; boundary contact is allowed, and interior points would count as covered.

Keep the final answer attached to its assumptions: exactly P, Q, and R, ordinary Euclidean distance, a closed filled disk, unrestricted real center coordinates, and a nonnegative radius. Under those conditions, C = (0, 3/4) with radius 5/4 is feasible and uniquely optimal. The squared-distance identity explains why every alternative center needs a larger covering radius, without turning a schematic illustration into a measurement.

Keep the three points with the disk that covers them.

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