About the Circle Equation Calculator
This circle equation calculator moves between the two ways a circle is written in algebra: standard (center-radius) form (x − h)² + (y − k)² = r², and general form x² + y² + Dx + Ey + F = 0. Enter a center and radius to get both equations, or enter the general-form coefficients to recover the center and radius — the step that usually needs completing the square.
It can also build the circle from other information: a center plus any point on the circle, or three points that the circle must pass through (the circumscribed circle of a triangle). That last option is useful beyond homework — for example to find the radius of a curved kerb, pipe or arch from three measured points.
Along with both equations you get the center coordinates, radius, diameter, circumference and area. If a general-form equation does not describe a real circle (the radius squared comes out zero or negative), the calculator tells you so instead of returning a meaningless answer.
How to use the circle equation calculator
- 1Choose what you are starting from.
- 2Enter the center and radius, the D, E and F coefficients, or the points.
- 3Read the standard-form equation (the primary result) and the general form.
- 4Use the center, radius, circumference and area for follow-up questions.
Formula and method
Every point on a circle is the same distance r from the center (h, k), and writing that distance with the distance formula gives the standard form (x − h)² + (y − k)² = r². Expanding the squares gives the general form, where D = −2h, E = −2k and F = h² + k² − r².
Going backwards is completing the square: h = −D/2, k = −E/2 and r² = D²/4 + E²/4 − F. If that r² is negative the equation has no real graph, and if it is zero the “circle” is a single point. For three points, the calculator substitutes each point into the general form, solves the resulting 3×3 linear system for D, E and F, then converts to center and radius. Collinear points have no solution.
- (h, k)
- Center of the circle
- r
- Radius
- D, E, F
- General-form coefficients
Worked examples
Center (2, −3), radius 4
Substitute h = 2, k = −3 and r² = 16. Expanding gives x² − 4x + 4 + y² + 6y + 9 = 16, i.e. x² + y² − 4x + 6y − 3 = 0.
General form x² + y² + 6x − 8y + 9 = 0
h = −6/2 = −3 and k = 8/2 = 4. Then r² = 9 + 16 − 9 = 16, so the radius is 4.
Circle through (0, 0), (4, 0) and (0, 6)
The three points form a right triangle, so the circle’s center is the midpoint of the hypotenuse, (2, 3), and the radius is √(2² + 3²) = √13 ≈ 3.606.
Center (1, 1) through the point (4, 5)
The radius is the distance from (1, 1) to (4, 5): √(3² + 4²) = 5, so r² = 25 and F = 1 + 1 − 25 = −23.
Frequently asked questions
What is the standard form of a circle equation?+
(x − h)² + (y − k)² = r², where (h, k) is the center and r the radius. A circle centered at the origin simplifies to x² + y² = r².
How do I find the center and radius from the general form?+
Complete the square, or use the shortcut h = −D/2, k = −E/2 and r = √(h² + k² − F). For x² + y² + 6x − 8y + 9 = 0 that gives center (−3, 4) and radius 4.
How do I write a circle equation from three points?+
Substitute each point into x² + y² + Dx + Ey + F = 0 and solve the three equations for D, E and F. The calculator does this automatically and converts the result to center-radius form.
Why does my equation say it is not a circle?+
If h² + k² − F is negative there is no real radius, so no points satisfy the equation. If it equals zero, the only solution is the single point (h, k).
What if the x² and y² coefficients are not 1?+
Divide the whole equation by that coefficient first. For 2x² + 2y² − 8x + 4y − 10 = 0, divide by 2 to get x² + y² − 4x + 2y − 5 = 0, then enter D = −4, E = 2, F = −5.