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Free Square in a Circle Calculator

Find the circle's radius from a square's side (or vice versa) for a square inscribed in a circle.

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When a square is inscribed in a circle (meaning all four of the square's corners touch the circle's edge), the square's diagonal is exactly equal to the circle's diameter: 2r = s√2. This calculator converts between the square's side length and the circle's radius using that relationship, and is a common building block in geometry problems, in mechanical design (fitting a square shaft or bracket within a circular housing), and in packing/layout problems where a square component must fit exactly within a circular boundary.

How it works

Enter either the square's side length or the circle's radius. The relationship comes directly from the Pythagorean theorem: the square's diagonal d satisfies d² = s² + s² = 2s², so d = s√2. Since that diagonal is also the circle's diameter (2r), we get 2r = s√2, which rearranges to r = s√2/2 (equivalently s/√2), or solved the other way, s = r√2 = 2r/√2. The calculator applies whichever direction you need and also reports the diagonal (equal to 2r) and the area of both shapes.

  1. Enter given value.
  2. Enter value.
  3. Click Calculate to see your results.

Examples

Side = 10

Radius = 10√2/2 = 5√2 ≈ 7.071, diagonal = 10√2 ≈ 14.142.

Radius = 5

Side = 5√2 ≈ 7.071, diagonal = 2 × 5 = 10.

Side = 6

Radius = 6√2/2 = 3√2 ≈ 4.243, diagonal = 6√2 ≈ 8.485.

Who should use it

  • Finding the largest square that fits inside a circular opening.
  • Geometry coursework on inscribed shapes.
  • Mechanical design problems fitting a square component within a circular boundary.

Industry applications

  • Geometry education
  • Mechanical and industrial design

Advantages

  • Converts in both directions between side length and radius.
  • Also reports diagonal and both shapes' areas in one step.

Limitations

  • Only covers the specific inscribed-square-in-circle configuration, not a circle inscribed in a square or other polygon-circle combinations.

Common mistakes to avoid

  • Confusing an inscribed square (corners touch the circle) with a circle inscribed in a square (circle touches the square's sides) — these describe different shapes and use different formulas.
  • Forgetting to account for the √2 factor and assuming the side length simply equals the radius or diameter.
  • Mixing up the diagonal (s√2) with the side length (s) when reading off results.

Best practices

  • Double check which configuration your problem actually describes — square-in-circle or circle-in-square — before choosing a formula.
  • Use the diagonal = diameter shortcut as a quick way to verify your answer by hand.
  • Remember the area ratio (2/π ≈ 63.7%) as a useful sanity check on computed areas.

Tips

  • Need the reverse configuration (a circle inscribed inside a square) instead? That uses diameter = side length directly, a simpler relationship without the √2 factor.

Frequently asked questions

Because all four corners of the inscribed square lie exactly on the circle, and the diagonal is the longest straight line connecting two of those corners — the two farthest-apart points on the circle are always exactly a diameter apart, and the square's diagonal happens to connect two such opposite points.
They are the reverse configuration of each other: here, the square's corners touch the circle (the circle "circumscribes" the square); in the other case, the circle touches the midpoints of the square's sides (the circle is inscribed inside the square), and that relationship instead uses the circle's diameter equal to the square's side length, not its diagonal.
The square's area is s², and the circle's area is πr² = π(s√2/2)² = πs²/2, so the square occupies exactly 2/π (about 63.7%) of the circle's area.
The 2r = s√2 relationship holds for any square inscribed in any circle — it's a fixed geometric ratio, not dependent on the specific size chosen.
That's exactly this calculation in the radius-to-side direction: s = r√2, giving the side length of the largest square whose corners all just touch the circle's boundary.

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