How this instrument works
Two endpoints of a diameter carry more information than two arbitrary points on a circle ever could, because a diameter is defined as a chord that runs straight through the center. That single constraint hands over the whole circle for free: the center has to sit exactly halfway along that chord — the ordinary midpoint of the two endpoints — and the radius has to be exactly half the chord's own length, since both given points sit on the circle's edge by definition. One segment, two facts, no extra equation needed.
Both outputs are familiar tools reused for a sharper purpose. Averaging the two endpoints' coordinates gives the center, and running the ordinary two-point distance formula between them, then halving it, gives the radius. What changes is not the arithmetic but what it is allowed to mean: applied to a random pair of points, an average is just an average, but applied specifically to a diameter, that same average is provably the circle's center, because a diameter by definition passes through — and is bisected by — that point.
The formulas only earn their answer when the two points genuinely sit at opposite ends of one diameter, not just anywhere on the circle. Hand it two points pulled from the same short stretch of arc instead, and the averaging step still spits out a number — it just lands well short of the true center, since a chord that skips over the center produces a midpoint that skips over it too. Push the input toward its other extreme and let both endpoints collapse onto one coordinate, and the radius shrinks to exactly zero — a diameter with no length left, the single point a circle degenerates into.
- Type the coordinates of one end of the diameter into Diameter endpoint 1: x and Diameter endpoint 1: y.
- Type the coordinates of the far end into Diameter endpoint 2: x and Diameter endpoint 2: y — any two points work, provided they truly mark opposite ends of one diameter.
- Center: x and Center: y report the circle's midpoint; Radius reports half the distance separating your two points.
- Try entering the two points in the reverse order — Center: x, Center: y, and Radius all come back unchanged.
- Assemble the three results yourself into (x − Center: x)² + (y − Center: y)² = Radius², the circle's complete equation.
Worked example — the diameter from (0, 0) to (6, 8)
Take the diameter running from (0, 0) to (6, 8). The center sits at the midpoint of those two points: cx = (0 + 6) ⁄ 2 = 3.0 and cy = (0 + 8) ⁄ 2 = 4.0, giving a center of (3, 4) without a single square root involved — averaging is all the center ever needs.
The radius needs the gap between the two endpoints, halved: √((6−0)² + (8−0)²) = √(36 + 64) = √100 = 10, so radius = 10 ⁄ 2 = 5.0 — the familiar 6-8-10 right triangle, itself the 3-4-5 triangle doubled. The full equation, (x − 3)² + (y − 4)² = 25, checks itself against both starting points: (0−3)² + (0−4)² = 9 + 16 = 25, and (6−3)² + (8−4)² = 9 + 16 = 25 — both original endpoints satisfy the very equation built from them.
Questions
What is the equation of a circle given the endpoints of its diameter?
It's (x − cx)² + (y − cy)² = r², where cx and cy are the midpoint of the two endpoints and r is half the distance between them. Averaging locates the center; halving the length between the points gives the radius — together they supply everything the standard center-radius form of a circle's equation needs.
How is a circle's center found from two diameter endpoints?
Average the coordinates: cx = (x₁ + x₂) ⁄ 2 and cy = (y₁ + y₂) ⁄ 2. This shortcut works only because a diameter is, by definition, a chord passing straight through the center, so the center has no choice but to sit exactly halfway between the two endpoints — an ordinary midpoint calculation, applied to a segment guaranteed to be bisected by the point you're after.
Why is the radius exactly half the distance between the two endpoints?
Because both endpoints lie on the circle itself, and a radius is by definition the length spanning the center out to a point on that boundary. Walking the full diameter — √((x₂−x₁)² + (y₂−y₁)²) — covers that same span twice over, once out to each endpoint through the shared center, so cutting the measured length in half hands back a single radius exactly.
Does it matter which endpoint I enter first?
No. Swapping Diameter endpoint 1 and Diameter endpoint 2 flips the sign of every coordinate difference inside the formulas, but averaging cancels a sign flip automatically, and the distance formula squares each difference before adding — both operations erase the sign, so Center: x, Center: y, and Radius come back identical whichever point you call first.
What if my two points aren't actually the ends of a diameter?
Then the result is wrong, even though the calculator still returns a number. Averaging only finds the true center when the segment it's given passes through that center — true of a diameter by definition, false of a general chord. Two arbitrary points on a circle need a third point, or that chord's own perpendicular bisector, to pin down the actual center instead.
Why not just apply the distance formula and the midpoint formula on their own?
This calculator applies the same two formulas, but reads them together as one geometric object rather than two isolated numbers: the midpoint becomes the circle's center, and half the distance becomes its radius, so the pair (cx, cy, r) directly builds the standard equation (x − cx)² + (y − cy)² = r² — a full circle, not just a point and a length sitting side by side.