Three points, two perpendicular bisectors, and their circumcenterLive current inputs
Math & Statistics
Circle Step-by-Step Calculator
Construct the unique circle through three non-collinear Cartesian points, exposing the determinant, circumcenter coordinates, radius, expanded equation, and point residual checks.
Circle center (h, k)-
Circle radius-
Standard equation-
Expanded equation-
Non-collinearity determinant-
Point A radial residual-
Point B radial residual-
Point C radial residual-
CONSTRUCTION SEQUENCE
Three points, two perpendicular bisectors, and their circumcenter
The coordinate drawing shows the entered points, connecting triangle, two midpoint bisectors, solved center, and the circle that reconciles all three radii.
CONSTRUCTION AUDIT
From chord midpoints to equal-radius residuals
Each row exposes a construction quantity or independent point-on-circle check.
| Step | Object | Symbolic relation | Computed value | Check |
|---|
POINT ENTRY
Use three distinct points from one coordinate system
- Enter points A, B, and C in consistent coordinates.
- Avoid duplicate or nearly collinear points.
- Inspect the determinant before trusting a very distant center.
- Read the graph for perpendicular-bisector geometry.
- Confirm all three radial residuals are near zero.
UNIQUENESS CONDITION
Three non-collinear points determine exactly one circle
If the points are collinear, their chord bisectors are parallel or coincident and no finite unique circle exists. A very small determinant creates an ill-conditioned construction.
The residuals are not a least-squares fit. For valid finite inputs, the algebraic circle is required to pass through all three points to floating-point precision.
DETERMINANT CIRCUMCENTER
Subtract squared-distance equations to obtain a linear center system
Equating the center-to-point squared distances cancels h² and k². Two linear equations solve the center; the radius then comes from any entered point.
Detailed calculation process and general formulas
D = 2[x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)]h = [q₁(y₂-y₃)+q₂(y₃-y₁)+q₃(y₁-y₂)]/Dk = [q₁(x₃-x₂)+q₂(x₁-x₃)+q₃(x₂-x₁)]/Dr = √[(x₁-h)²+(y₁-k)²](x-h)²+(y-k)²=r²Symbols, meanings, and units
- A,B,C
- three entered Cartesian pointscoordinate units
- qᵢ
- xᵢ²+yᵢ²coordinate units²
- D
- twice the oriented circumcenter determinantcoordinate units²
- h,k
- circle center coordinatescoordinate units
- r
- circle radiuscoordinate units
- εᵢ
- computed radius to point i minus rcoordinate units
CONSTRUCTION EVIDENCE
The center must satisfy both geometry and algebra
The page supplies three independent ways to challenge the result.
Determinant gate
-The determinant blocks division for collinear input.
Bisector intersection
-The visual center lies where two chord bisectors meet.
Equal radii
-All three radial residuals should reconcile to zero.
Decision takeaway: A huge radius from nearly collinear points can be mathematically correct but operationally unstable.
DATA QUALITY
Warning signs in point observations
- Repeated point
- Nearly straight three-point alignment
- Mixed coordinate frames
- Rounded map coordinates
- Axis swapped on one point
- Insufficient decimal precision
Applied decisions
Two three-point circle constructions
Arc reconstruction
Three measured points on a damaged circular edge are entered.
What the result clarifies: The center and radius reconstruct the nominal arc, subject to measurement noise.
Coordinate geometry exercise
Three exact vertices define a circumcircle.
What the result clarifies: The determinant, bisectors, and residuals provide a complete proof path.
Worked default scenario
Current-input substitution and reconciliation
Method references
References for this calculator's specific method
Scope and limitations
This page computes an exact circle through three entered points. It is not a best-fit circle for noisy point clouds. Nearly collinear coordinates can produce extreme centers and radii; use a conditioned least-squares fit for many measured points.
Circle Through Three Points Calculator | Step-by-Step Center and Equation FAQ
Why is there no solution for collinear points?
No finite circle can pass through three distinct points on one straight line.
Why is the radius extremely large?
The points are likely close to collinear, placing the perpendicular-bisector intersection far away.
What should the residuals be?
They should be near floating-point zero for a valid three-point construction.
Can this fit more than three points?
No. A noisy multi-point fit needs a different least-squares model.