Determine a Circle from Its Neighbors

Three fixed tangent neighbors

Begin with three circles: the unit enclosure and two equal circles inside it. Each inner radius is one half. They touch each other and the enclosure. Their signed curvatures are minus one, two, and two. The enclosure sign records its orientation, while all three physical radii remain positive.

Look at the gap below the two inner circles. A new circle must touch both of them and the enclosure. These are three constraints, not a request to place any convenient circle in the empty region. We seek a center and a radius that satisfy all three simultaneously.

Complete the tangent root. The lower circle has radius one third and curvature three. Its center lies two thirds of a unit below the enclosure center. The new circle meets all three original boundaries without crossing them. Center and contact guides make those relationships visible.

Return to the complete root of four mutually tangent circles. Mutually tangent means every pair in this set touches. The signed curvature values are minus one, two, two, and three. We will check both the algebraic relation and the geometry of the new circle.

Check the curvature relation

For four mutually tangent circles in this configuration, the square of the sum of their signed curvatures equals twice the sum of their squared curvatures. This is the Descartes circle relation. The enclosure’s negative sign participates in the sum before it is squared.

Add the four displayed signed values: minus one plus two plus two plus three equals six. The square of that sum is thirty-six. Do not replace minus one by positive one; doing so would describe a different oriented configuration and would fail this calculation.

Now square each curvature separately. One, four, four, and nine sum to eighteen. Twice eighteen is thirty-six, matching the other side. Notice that the negative value contributes positive one when squared, but contributed negative one to the earlier signed sum.

This check confirms the measurements are consistent with the circle relation. It does not by itself draw the centers. The World also uses the geometric tangency conditions to locate the circle. Curvature is one number; a planar circle additionally needs the two coordinates of its center.

Verify the actual tangencies

Approach the lower circle. Its center is at zero, minus two thirds, and its radius is one third. Its center is two thirds from the enclosure center. Add its radius and that center distance: one third plus two thirds is one, the enclosure radius.

Compare the right half-radius circle. Its center is at one half, zero. The horizontal center difference is one half and the vertical difference is two thirds. By the Pythagorean theorem, the center distance is the square root of one quarter plus four ninths: five sixths.

The two positive radii also sum to five sixths: one half plus one third. Equality between center distance and radius sum proves external tangency for this pair. The left pair has the same distances by reflection across the vertical axis.

Return to the root. We have checked internal tangency to the enclosure and external tangency to both inner neighbors. Algebra and geometry now describe the same object. A curvature equation without the center conditions would leave part of that description unfinished.

Two locations can have the same curvature

Keep the original three circles fixed and add the upper tangent solution. The upper and lower circles both have radius one third and curvature three. They are different circles because their centers are different. Equal curvature does not imply equal position.

Move toward the upper solution. Its center is at zero, positive two thirds. It touches the same three original neighbors as the lower solution. The two solutions are reflections across the horizontal axis, which leaves those original three circles unchanged.

Return to the lower solution while keeping both visible. Its center has negative two thirds as its vertical coordinate. The two equal curvature values form a special symmetric case. In a less symmetric set of three tangent neighbors, the two solution curvatures need not be equal.

The replacement relation predicts the other curvature: twice the sum of the three fixed bends, minus the old bend. Here twice minus one plus two plus two, then subtract three, gives three. The repeated answer is consistent with two distinct positions.

Hold three inner circles fixed

Now change which three circles remain fixed. Keep the two half-radius circles and the lower third-radius circle. Their curvatures are two, two, and three. One tangent solution is the unit enclosure with curvature minus one. The other is a much smaller circle in their central gap.

The replacement rule gives twice two plus two plus three, minus negative one. Twice seven is fourteen; subtracting negative one adds one. The new curvature is fifteen, so the new radius is one fifteenth. The native selection shows that small circle and its three determining parents.

This circle is centered at zero, minus four fifteenths. It lies between the two half-radius circles and the lower third-radius circle. The enclosure is not one of its three determining parents in this replacement. It is the alternative solution for that same parent set.

Reverse the calculation. Twice seven minus fifteen is minus one, recovering the enclosure’s signed curvature. A replacement is reversible when the three neighbors remain fixed. The negative result is meaningful here because it identifies an enclosing solution, not a negative physical length.

Predict a different replacement

Choose another set of three neighbors: the enclosure, the left half-radius circle, and the lower third-radius circle. Their signed curvatures are minus one, two, and three. The original right half-radius circle has curvature two. Predict its replacement before reading the selected result.

Twice the sum of minus one, two, and three is eight. Subtract the old curvature two to obtain six. The newly selected left-side circle therefore has radius one sixth. Its center is at minus one half, minus two thirds.

The circle of curvature six is larger than the circle of curvature fifteen. Their radii are one sixth and one fifteenth, respectively. Both arose from the same replacement rule, but different parent sets supplied different sums. The local neighbors determine which gap is being filled.

Check reversibility once more. Twice four minus six equals two. That returns the replaced half-radius circle’s curvature. The calculation does not create a new rule for each gap; it applies one relation to the specific three fixed neighbors of that gap.

Verify both symmetric solutions with center distances

Fix the unit enclosure and the two radius-one-half circles. Their centers are one half unit left and right of the enclosure center. The lower tangent circle has center zero, negative two thirds and radius one third. We can verify its contacts using these actual positions, independently of its curvature label.

The distance from the lower center to either half-radius center is the square root of one quarter plus four ninths. That is five sixths, exactly one half plus one third. Both external contacts therefore satisfy the center-distance condition. The symmetric horizontal offsets produce equal distances to the two neighbors.

Its center is two thirds from the enclosure center. Add the new radius one third to get one, the enclosure radius. This verifies internal tangency to the enclosing boundary. The same circle therefore satisfies two external contacts and one internal contact, with the distinction determined by containment.

Now inspect the upper symmetric solution at center zero, positive two thirds. It also has radius one third and curvature three. Its distances to the fixed neighbors are unchanged by reflection. Equal curvature does not identify a unique circle unless the position or branch of the geometric solution is also specified.

Return to the whole configuration with both solutions retained. The upper and lower circles are separate geometric objects even though both carry the same bend. This is why the World’s selection retains center and object identity as well as a numerical curvature: a number alone cannot tell us which tangent circle is being discussed.

A determined circle has size and position

Return to the initial root. Three mutually tangent neighbors constrained the fourth circle. The signed curvature relation checked its reciprocal radius, while distances between centers checked its placement. Both types of information were needed to describe the tangent configuration.

Show the two symmetric third-radius solutions again. Their curvatures coincide, but their centers do not. This example prevents a common shortcut: solving for a curvature is not the same as uniquely identifying a geometric circle in the plane.

Show the first generation of replacements. Each selected new circle keeps three neighbors fixed and replaces the fourth member of a tangent quadruple. Repeating this local operation will fill more gaps, which is the subject of the next lesson.

Read the complete pattern once more. For circles, the replacement bend is twice the sum of three fixed bends minus the old bend. Keep signed values, distinguish radius from curvature, and identify the actual three neighbors. Those habits make the recursive construction traceable.