Circle Equation Calculator
Enter the centre (h, k) and radius r, and get the circle’s equation in standard form (x−h)²+(y−k)²=r² and expanded general form x²+y²+Dx+Ey+F=0, with every coefficient shown.
Standard form
(x − 2)² + (y + 3)² = 25
- General form
- x² + y² − 4x + 6y − 12 = 0
- Coefficients (D, E, F)
- D = -4, E = 6, F = -12
- Centre
- (2, -3)
- Radius
- 5
- Diameter
- 10
- Circumference (2πr)
- 31.42
- Area (πr²)
- 78.54
Every point (x, y) on this circle is exactly 5 units from the centre (2, -3) — the standard form states that with the distance formula, and expanding it gives the general form.
How to use this calculator
Type the centre’s x-coordinate (h), the centre’s y-coordinate (k), and the radius (r). Either coordinate can be negative, zero, or a decimal; the radius must be positive. The calculator writes out the standard form with the signs resolved — so a centre of (2, −3) correctly appears as (x − 2)² + (y + 3)² — then expands it into the general form and lists the coefficients D, E and F separately, along with the circle’s diameter, circumference and area.
How the calculation works
A circle is the set of all points at distance r from the centre (h, k). Applying the distance formula to a general point (x, y) and squaring both sides gives the standard form: (x − h)² + (y − k)² = r². Expanding the two squares and collecting terms produces the general form x² + y² + Dx + Ey + F = 0, where D = −2h, E = −2k and F = h² + k² − r². The two forms describe exactly the same circle — the standard form makes the centre and radius visible at a glance, while the general form is the shape you meet when an equation arrives already multiplied out. To go back the other way, complete the square: the centre is (−D/2, −E/2) and r = √(D²/4 + E²/4 − F).
Worked example
Take a circle with centre (2, −3) and radius 5. Standard form: (x − 2)² + (y + 3)² = 25 — note the sign flips, because subtracting a negative k gives y + 3. Expanding: x² − 4x + 4 + y² + 6y + 9 = 25, and moving 25 across gives the general form x² + y² − 4x + 6y − 12 = 0, so D = −4, E = 6, F = −12. As a check, the point (6, 0) should lie on this circle: (6 − 2)² + (0 + 3)² = 16 + 9 = 25 ✓. The diameter is 10, the circumference 2π×5 ≈ 31.42, and the area π×25 ≈ 78.54.
Frequently asked questions
What is the equation of a circle?
In standard form it is (x − h)² + (y − k)² = r², where (h, k) is the centre and r is the radius. It comes straight from the distance formula: a point (x, y) is on the circle exactly when its distance from the centre equals r, and squaring that condition removes the square root. A circle centred at the origin simplifies to x² + y² = r².
What is the general form of a circle equation?
x² + y² + Dx + Ey + F = 0. It is the standard form multiplied out: expanding (x − h)² + (y − k)² = r² gives D = −2h, E = −2k and F = h² + k² − r². Both forms describe the same circle; the general form just hides the centre and radius inside its coefficients. Note the x² and y² terms always have coefficient 1 — if your equation has Ax² + Ay² with A ≠ 1, divide everything by A first.
How do I find the centre and radius from the general form?
Complete the square, or read them from the coefficients directly: the centre is (−D/2, −E/2) and the radius is √(D²/4 + E²/4 − F). For example, x² + y² − 4x + 6y − 12 = 0 has D = −4, E = 6, F = −12, so the centre is (2, −3) and r = √(4 + 9 + 12) = √25 = 5.
Why does (x − 2)² + (y + 3)² = 25 have a centre of (2, −3) and not (2, 3)?
The standard form subtracts the centre coordinates: (x − h)² + (y − k)². When k = −3, subtracting it gives y − (−3) = y + 3, so a plus sign inside the bracket means a negative coordinate. The sign flip trips a lot of people up — the centre is always the value that makes each bracket zero.
What if D²/4 + E²/4 − F is zero or negative?
Then the general-form equation does not describe a real circle. If the expression is exactly zero the radius is zero and the “circle” is the single point (−D/2, −E/2), sometimes called a degenerate or point circle. If it is negative there is no real solution at all — no point (x, y) satisfies the equation. This calculator works from a positive radius, so it always produces a genuine circle.
How do I tell whether a point is inside, on, or outside the circle?
Substitute the point into the left side of the standard form and compare with r². If (x − h)² + (y − k)² is less than r² the point is inside; equal means it lies on the circle; greater means it is outside. For the default circle, the origin gives (0 − 2)² + (0 + 3)² = 13 < 25, so the origin is inside.