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Area of Crescent (Lune) Calculator

A crescent, also called a lune, is the region of one circle that lies outside a second overlapping circle. Enter the two radii and the distance between centres and this calculator instantly returns the area of each lune, the shared lens (intersection) area, the perimeter of the crescent arc, and a full step-by-step breakdown of the arithmetic. Switch between metric and imperial units as needed.

Your details

Radius of the larger (outer) circle. Must be greater than 0.
cm
Radius of the second (inner) circle. The two circles must overlap.
cm
Centre-to-centre distance. Must be between |r1 - r2| and r1 + r2 for the circles to overlap without one containing the other.
cm
Lune 1 area
142.2597

Area of circle 1 outside circle 2 (the main crescent)

Lune 2 area54.2951
Lens (overlap) area58.8022
Circle 1 total area201.0619
Circle 2 total area113.0973
Crescent perimeter (arc)53.0788
_unitLabelcm
Lune 1142.2597
Lens (overlap)58.8022
Lune 254.2951

Lune 1 area: 142.2597 cm². Lune 2 area: 54.2951 cm².

  • Lune 1 covers 70.8% of circle 1's total area of 201.0619 cm².
  • The shared lens region is 58.8022 cm², representing the overlap of both circles.
  • Lune 1 is larger than lune 2, meaning circle 1 extends further beyond the overlap than circle 2 does.
  • A crescent is a lune whose inner arc does not pass through the centre of the outer circle.

Next stepTo confirm the result, add the two lune areas and the lens area together: the sum should equal the sum of both full circle areas.

Formula

lens=r12α+r22βr12sinαcosαr22sinβcosβ,α=arccos ⁣(d2+r12r222dr1),β=arccos ⁣(d2+r22r122dr2)\text{lens} = r_1^2 \alpha + r_2^2 \beta - r_1^2 \sin\alpha\cos\alpha - r_2^2 \sin\beta\cos\beta, \quad \alpha = \arccos\!\left(\tfrac{d^2+r_1^2-r_2^2}{2dr_1}\right), \quad \beta = \arccos\!\left(\tfrac{d^2+r_2^2-r_1^2}{2dr_2}\right)

Worked example

For r1 = 8 cm, r2 = 6 cm, d = 7 cm: alpha = arccos((49 + 64 - 36) / (2 x 7 x 8)) = arccos(77/112) approx 0.8122 rad. beta = arccos((49 + 36 - 64) / (2 x 7 x 6)) = arccos(21/84) = arccos(0.25) approx 1.3181 rad. Lens = 8^2 x 0.8122 + 6^2 x 1.3181 - 8^2 x sin(0.8122)cos(0.8122) - 6^2 x sin(1.3181)cos(1.3181) approx 51.98 + 47.45 - 46.00 - 8.42 approx 45.01 cm^2. Lune 1 = pi x 64 - 45.01 approx 201.06 - 45.01 = 156.05 cm^2. Lune 2 = pi x 36 - 45.01 approx 113.10 - 45.01 = 68.09 cm^2.

What is a crescent (lune)?

In geometry, a lune is the region enclosed between two intersecting circular arcs. When the inner arc does not pass through the centre of the outer circle, the resulting shape is specifically called a crescent. You encounter crescent shapes constantly: the Moon in its various phases, logos and heraldic symbols, architectural arches, and decorative motifs across many cultures all draw on this elegant form. Mathematically, the crescent is simply the area of one circle minus the area where that circle overlaps a second circle, a quantity sometimes called the "lens" or "vesica" region.

How to use this calculator

Enter the radius of each circle (r1 and r2) and the straight-line distance between their centres (d). The distance must satisfy |r1 - r2| < d < r1 + r2, otherwise the circles either fail to overlap at all, or one sits completely inside the other and no crescent exists. The calculator returns: the area of each lune, the lens (intersection) area, the complete area of each full circle, and the perimeter of the main crescent arc. All values update live as you type. Switch between metric and imperial units with the unit selector at the top; both radii and the centre distance change their label to match.

The lens area formula explained

The lens area is computed from two circular segments. Each segment is the region between a chord and the arc it cuts off. First, find the half-angle alpha at centre O1 using the law of cosines: alpha = arccos((d^2 + r1^2 - r2^2) / (2 x d x r1)). Similarly, beta = arccos((d^2 + r2^2 - r1^2) / (2 x d x r2)). The lens area then equals r1^2 x alpha + r2^2 x beta minus the two triangular areas r1^2 x sin(alpha)cos(alpha) and r2^2 x sin(beta)cos(beta). Subtracting this lens from pi x r1^2 gives lune 1, and subtracting it from pi x r2^2 gives lune 2. The "show your work" panel below the result card walks through each step with your actual numbers.

Special case: two circles with the same radius

When r1 equals r2, the configuration is symmetric and both lune areas are identical. The lens becomes a vesica piscis, a shape with a width-to-height ratio of 1:1.732. For equal radii the formula simplifies: lens = r^2 x (2 x arccos(d / (2r)) - sin(2 x arccos(d / (2r)))). At d = 0 the circles coincide and the lens fills each circle entirely; as d approaches 2r the circles barely touch and the lens area shrinks to zero. Architects and designers often work with equal-radius crescents because of the inherent bilateral symmetry.

Real-world applications

Crescent geometry appears in a surprising variety of fields. Naval architects use lune areas when modelling waterplane sections of ships. Lens designers calculate the overlap between two spherical surfaces in exactly this way. In GIS and mapping software, the intersection of two circular coverage zones (such as two cell-tower ranges) is a lens, and the non-overlapping portions are lunes. Astronomers compute the illuminated fraction of the Moon's visible disc using a closely related formula. Urban planners estimating the shared service area of two circular parks also apply the same maths. Knowing how to decompose the overlap into lens plus lunes is a transferable skill across all of these domains.

Crescent geometry - key relationships

QuantityFormulaNotes
Circle 1 areaπ r1²Full disc including lens
Circle 2 areaπ r2²Full disc including lens
Lens arear1² α + r2² β - r1² sinα cosα - r2² sinβ cosβα and β from law of cosines
Lune 1 areaπ r1² - lensRegion of circle 1 outside circle 2
Lune 2 areaπ r2² - lensRegion of circle 2 outside circle 1
Lune 1 + Lune 2 + Lensπ r1² + π r2² - lensCheck: union of both circles
Valid range for d|r1 - r2| < d < r1 + r2For a proper crescent (no containment)

Relationships that always hold for two overlapping circles forming a crescent, regardless of radii or centre distance.

Frequently asked questions

What is the difference between a lune and a crescent in geometry?

A lune is any region bounded by two circular arcs. A crescent is a specific type of lune: it forms when the inner arc of the lune does not pass through the centre of the outer circle, producing the pointed, Moon-like shape familiar in everyday language. In common usage the two terms are often interchangeable, but in strict geometry a crescent is a proper subset of lunes.

What values of d produce a valid crescent?

The centre-to-centre distance d must be strictly between |r1 - r2| and r1 + r2. If d is at or above r1 + r2, the circles touch externally or do not meet, so no overlap exists. If d is at or below |r1 - r2|, one circle sits completely inside the other and no crescent forms. Only when the circles genuinely overlap - without either containing the other - do you get two distinct lunes.

How do I calculate the area of a simple Moon crescent?

Model the illuminated crescent as lune 1: the region of the large (sunlit) circle that lies outside a slightly smaller or offset shadow circle. Enter the radius of the Moon disc as r1, the radius of the shadow circle as r2, and the horizontal offset of the shadow as d. Lune 1 is then the crescent area. Because the Moon's tilt changes over time you may need to rotate the result, but the area formula is the same.

Can both lune areas be equal?

Yes, but only when both circles have the same radius (r1 = r2). In that case the configuration is symmetric: the lens lies midway between the centres, and each lune is a mirror image of the other, so both lune areas are identical. When the radii differ, the lune of the larger circle is always the bigger region.

What is the perimeter of lune 1?

The boundary of lune 1 consists of two arcs: the major (outer) arc of circle 1, which spans the reflex angle (2 pi - 2 alpha) at centre O1, and the minor arc of circle 2, which spans angle 2 beta at centre O2. The total perimeter is therefore (2 pi - 2 alpha) x r1 plus 2 beta x r2. This calculator computes and displays that value.

Is the area formula affected by whether I use metric or imperial units?

No, the formula is the same in any consistent unit system. Just make sure all three inputs (r1, r2, d) are in the same unit. The calculator accepts centimetres and metres (metric) or inches, feet, and yards (imperial). The area result will then be in the corresponding squared unit (cm^2, in^2, and so on).

Sources

Written by Dr. Elena Vasquez, PhD Mathematician · Lisbon, Portugal

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