Last updated: 2026-07-01

The equation of a circle

A circle is the set of all points at a fixed distance — the radius $r$ — from a fixed point — the center $(h, k)$. Translating that definition into coordinates gives the standard form of a circle:

$(x - h)^2 + (y - k)^2 = r^2$

Multiplying out the squares and collecting terms produces the general form:

$x^2 + y^2 + Dx + Ey + F = 0$

Both equations describe the same circle. The standard form puts the center and radius on display; the general form hides them but is the shape that naturally appears when an equation is given to you "raw". This calculator converts between the two in either direction.

From general form to center and radius

Given $x^2 + y^2 + Dx + Ey + F = 0$, the center and radius are:

$\text{center} = \left(-\frac{D}{2},\ -\frac{E}{2}\right) \qquad r = \sqrt{\frac{D^2}{4} + \frac{E^2}{4} - F}$

The center coordinates come straight from the coefficients of $x$ and $y$. The radius requires the quantity under the square root — the radicand $\tfrac{D^2}{4} + \tfrac{E^2}{4} - F$ — to be positive for a genuine circle to exist.

Completing the square

The conversion is just completing the square applied to both variables at once. Start from the general form and group:

$(x^2 + Dx) + (y^2 + Ey) = -F$

To complete each square, add the square of half the linear coefficient to both sides — $\left(\tfrac{D}{2}\right)^2$ for the $x$ group and $\left(\tfrac{E}{2}\right)^2$ for the $y$ group:

$\left(x + \tfrac{D}{2}\right)^2 + \left(y + \tfrac{E}{2}\right)^2 = \frac{D^2}{4} + \frac{E^2}{4} - F$

The left side is now a sum of two perfect squares centered at $\left(-\tfrac{D}{2}, -\tfrac{E}{2}\right)$, and the right side is $r^2$. That is exactly the standard form.

Worked example

Problem: Find the center and radius of $x^2 + y^2 - 4x + 6y + 9 = 0$.

Step 1 — identify the coefficients: $D = -4$, $E = 6$, $F = 9$.

Step 2 — find the center:

$\text{center} = \left(-\frac{-4}{2},\ -\frac{6}{2}\right) = (2,\ -3)$

Step 3 — find the radius:

$r = \sqrt{\frac{(-4)^2}{4} + \frac{6^2}{4} - 9} = \sqrt{4 + 9 - 9} = \sqrt{4} = 2$

So the circle has center $(2, -3)$ and radius $2$, and its standard form is $(x - 2)^2 + (y + 3)^2 = 4$.

From center and radius back to general form

Going the other way, expand the squares of the standard form. With center $(h, k)$ and radius $r$:

$D = -2h \qquad E = -2k \qquad F = h^2 + k^2 - r^2$

For the example above, $h = 2$, $k = -3$, $r = 2$ gives $D = -4$, $E = 6$, and $F = 4 + 9 - 4 = 9$, recovering the original general form $x^2 + y^2 - 4x + 6y + 9 = 0$.

When there is no real circle

The radius is only real when the radicand is positive:

$\frac{D^2}{4} + \frac{E^2}{4} - F > 0$

Three cases arise:

• Positive radicand — a real circle with $r = \sqrt{\text{radicand}}$.
• Zero radicand — the equation collapses to the single point $\left(-\tfrac{D}{2}, -\tfrac{E}{2}\right)$, a degenerate "point circle" of radius $0$.
• Negative radicand — no real point satisfies the equation. This is sometimes called an imaginary circle; geometrically there is nothing to draw.

For instance, $x^2 + y^2 + 1 = 0$ has $D = E = 0$ and $F = 1$, so the radicand is $-1$. No real $(x, y)$ can make a sum of squares equal $-1$, so there is no circle.

Why the general form matters

The general form $x^2 + y^2 + Dx + Ey + F = 0$ is the second-degree equation in which the $x^2$ and $y^2$ coefficients are equal and there is no $xy$ term — the algebraic signature of a circle among the conic sections. Recognising it lets you tell at a glance whether a messy second-degree equation is a circle (equal squared coefficients, no cross term) before doing any work, and completing the square then exposes its center and radius.

Frequently Asked Questions (FAQ)

What is the difference between the standard and general form of a circle?

The standard form (x − h)² + (y − k)² = r² shows the center (h, k) and radius r directly — you can read them straight off the equation. The general form x² + y² + Dx + Ey + F = 0 is the expanded version; it hides the center and radius but is convenient for algebra and for comparing with other conic sections.

The two forms describe the same circle, and you can convert either way by completing the square (general → standard) or expanding the squares (standard → general).

How do I find the center and radius from x² + y² + Dx + Ey + F = 0?

Complete the square separately for x and y. The center is (−D/2, −E/2), and the radius is r = √(D²/4 + E²/4 − F). For example, x² + y² − 4x + 6y + 9 = 0 has center (2, −3) and radius √(4 + 9 − 9) = √4 = 2.

How does completing the square turn the general form into the standard form?

Group the x-terms and the y-terms: (x² + Dx) + (y² + Ey) = −F. Add (D/2)² and (E/2)² to both sides to complete each square: (x + D/2)² + (y + E/2)² = D²/4 + E²/4 − F. The left side is now two perfect squares centered at (−D/2, −E/2), and the right side is r². This is the standard form.

When does the equation not represent a real circle?

The radius squared is r² = D²/4 + E²/4 − F. If this value is negative, no real (x, y) satisfies the equation, so there is no circle (sometimes called an imaginary circle). If it is exactly zero, the equation collapses to a single point at the center — a degenerate "point circle" of radius 0. A genuine circle requires D²/4 + E²/4 − F > 0.