How this instrument works
Any three points that don't fall on a single straight line sit on exactly one circle, and the reason comes from a single idea: a circle's center is the one location equidistant from every point on its edge. That means the center must be equally far from point 1 and point 2, and also equally far from point 2 and point 3. The set of locations equally distant from two given points is the perpendicular bisector of the segment joining them, so the center is wherever the perpendicular bisector of the first pair crosses the perpendicular bisector of the second pair.
Turning that geometric idea into numbers gives two linear equations in the two unknowns h and k, the center's coordinates — a 2-by-2 system solved here with Cramer's Rule, the same determinant-ratio technique used for ordinary simultaneous equations elsewhere on this site. Once h and k are known, the radius follows immediately as the plain distance from the center to any one of the three original points: r = √((x₁−h)² + (y₁−k)²).
This is a distinctly harder problem than converting a circle's general-form equation into standard form, which this site also covers. That page already starts from known coefficients D, E, and F describing an equation someone has already written down, and simply completes the square to read off the center and radius hiding inside it. Here, no equation exists yet at all — the circle itself has to be produced from nothing but three raw coordinate pairs, using the equidistant-center property to build the linear system that pins it down.
Three points that happen to sit on one straight line break the method entirely: their perpendicular bisectors run parallel to each other and never meet, so no finite center exists and the 2-by-2 system has no solution. This isn't a flaw in the approach — it reflects a genuine fact about geometry, since no circle can pass through three collinear points in the first place.
- Enter the first point's coordinates into Point 1: x and Point 1: y.
- Enter the second point into Point 2: x and Point 2: y, and the third into Point 3: x and Point 3: y.
- Read Center: x (h) and Center: y (k): the sheet solves the perpendicular-bisector system automatically.
- Read Radius: the distance from the recovered center back out to any of the three original points.
- Choose three points that aren't all on one line — collinear points describe no circle at all.
Worked example — points (0,0), (4,0), and (0,4)
Three points, (0,0), (4,0), and (0,4), go into Point 1 through Point 3. The perpendicular bisector of (0,0) and (4,0) is the vertical line x = 2; the perpendicular bisector of (0,0) and (0,4) is the horizontal line y = 2. Those two lines cross at exactly one location, (2,2) — no other point is equally far from all three inputs — so Center: x (h) reads 2 and Center: y (k) reads 2.
The radius follows as the distance from that center back to point 1: r = √((0−2)² + (0−2)²) = √8 = 2.8284271247461903, matching 2√2 exactly. Checking against point 2 confirms the same figure: √((4−2)² + (0−2)²) = √(4+4) = √8, identical to the first check, which is exactly what having found the true center guarantees.
Questions
What information do I need to find a circle's equation from scratch?
Three points that lie on the circle and aren't all on one straight line — nothing else. Given just those three coordinate pairs, this calculator reconstructs the center (h,k) and radius r that a circle through all three would have to carry, without any prior equation or known coefficients to start from.
Why does a perpendicular bisector find the circle's center?
Because every point on a perpendicular bisector of a segment is, by definition, equally distant from that segment's two endpoints. A circle's center must be equally distant from every point on the circle, so it has to sit on the perpendicular bisector of any chord — including the chord between any two of the three given points. Two such bisectors, from two different pairs, cross at exactly one location: the center.
How is this different from the general-to-standard-form circle page?
That page already has a circle's equation in hand, written as x² + y² + Dx + Ey + F = 0, and simply completes the square to expose the center and radius sitting inside those known coefficients. This page starts with no equation and no coefficients at all — just three raw points — and has to build the circle from the ground up using the equidistant-center property before any center or radius exists to read off.
What happens if the three points are collinear (all on one line)?
No circle passes through three points on a single straight line, and the math reflects that directly: the two perpendicular bisectors turn out parallel and never intersect, so the 2-by-2 system has no solution. This calculator can't return a center or radius for that input, because no genuine one exists.
Where does Cramer's Rule come into this calculation?
Once the equidistant-center condition is written out algebraically for two pairs of points, the result is two linear equations in the two unknowns h and k. That's exactly the shape Cramer's Rule is built to solve: a 2-by-2 system reduced to two ratio-of-determinants formulas, giving h and k directly rather than through substitution or elimination.