What this calculator does
A circle equation written as x² + y² + Dx + Ey + F = 0 hides its centre and radius. Completing the square in both variables recovers them, which is the whole point of the conversion.
The arithmetic is short. The centre is minus D over two and minus E over two, and the radius squared is whatever remains. For D = −4, E = 6 and F = 4 the centre is (2, −3) with a radius of exactly 3.000.
The formula
Half of each linear coefficient, negated, gives the centre coordinate. The radius squared is the sum of the squared centre coordinates less the constant term.
| Term | Meaning |
|---|---|
| General form | x² + y² + Dx + Ey + F = 0, which obscures the centre and radius. |
| Standard form | (x − h)² + (y − k)² = r², which states them directly. |
| Completing the square | The algebraic step that converts between the two. |
The inputs explained
| Field | What to enter |
|---|---|
| D (coefficient of x) | Coefficient of x in the general form. |
| E (coefficient of y) | Coefficient of y in the general form. |
| F (constant) | The constant term. A large positive value can make the radius imaginary. |
When to use it
Reading a circle equation
General form tells you nothing until it is converted.
Graphing a circle
Plotting requires the centre and radius, not the expanded form.
Checking whether an equation is a circle at all
A negative radius squared means no real circle exists.
Worked examples
Every figure in the tables below is produced by this page’s own calculator at build time, so the numbers and the tool always agree. Select any row to load that scenario.
How does the constant change the radius?
The same centre with three constant terms.
Questions
What if the radius squared comes out negative?
Then no real circle exists. The equation has no real solutions, and it is sometimes called an imaginary circle. The calculator reports this rather than attempting a square root of a negative number.
What if the radius squared is exactly zero?
The equation describes a single point, the centre itself, sometimes called a degenerate circle. It is the boundary case between a real circle and an imaginary one.
Why is the centre negative half the coefficient?
Because expanding (x − h)² gives x² − 2hx + h², so the x coefficient is −2h. Reversing that gives h as minus the coefficient over two, and the same applies to y.
What if the equation has an x² coefficient other than 1?
Divide the whole equation through by it first. If the x² and y² coefficients differ, the shape is an ellipse rather than a circle and this conversion does not apply.
For where a line meets the circle, see the circle line intersection calculator. For circle measurements, see the circle calculator.