How this instrument works
The equation x² + y² + Dx + Ey + F = 0 is what a circle's equation looks like after someone has already expanded it — multiplied out (x − a)² + (y − b)² = r² and collected the squared terms together with the constants. Once that expansion has happened, the center (a, b) and radius r are still fully determined, just buried inside D, E, and F rather than sitting in plain sight. Reversing the expansion — completing the square on the x terms and again on the y terms — is exactly how this sheet extracts them.
The recovery is direct because expanding (x − a)² gives x² − 2ax + a², so matching it to x² + Dx forces D = −2a, meaning a = −D ⁄ 2; the same reasoning on the y terms gives b = −E ⁄ 2. What's left over, F, absorbs a² + b² − r², so subtracting it back out and taking a square root hands back the radius: r = √((D⁄2)² + (E⁄2)² − F). Three coefficients, three geometric facts, recovered by running the algebra backwards.
Not every D, E, F triple describes a real circle. When (D⁄2)² + (E⁄2)² − F works out to exactly zero, the 'circle' has shrunk to a single point — D = E = −2 and F = 2 gives the equation x² + y² − 2x − 2y + 2 = 0, which is satisfied only by the point (1, 1). Push F a little further past that threshold and the quantity under the square root turns negative: the equation still balances algebraically, but no real (x, y) pair satisfies it, an imaginary circle in exactly the sense an imaginary root belongs to a quadratic with a negative discriminant.
- Read the expanded circle equation and enter the coefficient in front of x into the D field.
- Enter the coefficient in front of y into the E field.
- Enter the constant term — everything with no x or y attached — into the F field.
- Center: x and Center: y give the circle's center; Radius gives its size, already square-rooted.
- If the coefficients describe no real circle, the sheet's check flags it instead of returning a radius.
Worked example — center and radius from x² + y² − 4x − 6y + 4 = 0
Take the expanded equation x² + y² − 4x − 6y + 4 = 0, so D = −4, E = −6, and F = 4. The center's x-coordinate is cx = −D ⁄ 2 = −(−4) ⁄ 2 = 2, and cy = −E ⁄ 2 = −(−6) ⁄ 2 = 3 — the double negative in each is the whole trick, and it is exactly where a rushed calculation goes wrong. The radius is r = √((−4⁄2)² + (−6⁄2)² − 4) = √(4 + 9 − 4) = √9 = 3.
Expanding the standard form back out confirms it: (x − 2)² + (y − 3)² = 9 becomes x² − 4x + 4 + y² − 6y + 9 = 9, which simplifies to x² + y² − 4x − 6y + 4 = 0 — the exact equation this example started from. That round trip is the whole check: center (2, 3), radius 3, and the original coefficients rebuilt with nothing left over.
Questions
How do you find the center of a circle from its expanded equation?
Divide the coefficient of x by −2 to get cx, and the coefficient of y by −2 to get cy: cx = −D ⁄ 2, cy = −E ⁄ 2. Both come from completing the square on x² + Dx and y² + Ey separately, the same move used to solve a quadratic by completing the square rather than factoring.
Why is there a minus sign in cx = −D ⁄ 2 when D can already be negative?
Because expanding (x − a)² gives x² − 2ax, matching it to x² + Dx forces D = −2a, so a = −D ⁄ 2 — one negative sign is built into the formula itself. Plug in a negative D, like −4, and the two negatives cancel: cx = −(−4) ⁄ 2 = 2, not −2. Dropping that first negative is the single most common slip with this formula.
What does a negative value under the square root mean here?
It means the D, E, F triple does not describe a real circle at all. The algebra still balances — F still equals a² + b² − r² for some numbers — but r² comes out negative, which has no real square root, the same way a quadratic with a negative discriminant has no real roots. This sheet's check catches that case before returning a result.
How is this different from converting between a circle's radius, diameter, and circumference?
Those conversions assume you already know you have a circle and just need a different one of its measurements. This calculator starts one step earlier, with a raw expanded equation that merely looks like a circle, and has to determine the center and radius — and confirm a real circle even exists — before any of those familiar measurements can be taken.
What does it mean if F is exactly zero?
The circle passes through the origin. Setting x = 0 and y = 0 in x² + y² + Dx + Ey + F = 0 leaves only F, so the origin satisfies the equation precisely when F = 0 — a quick check worth remembering whenever F drops out of a derivation.
Why must the x² and y² coefficients both equal 1 before reading off D, E, and F?
This form assumes the equation has already been normalized. If the original equation reads A x² + A y² + Dx + Ey + F = 0 with A not equal to 1, divide every term by A first — otherwise the numbers you enter are not the D and E this formula expects, and the recovered center and radius come out wrong by a factor of A.