Circle Equation Calculator

What Is the Equation of a Circle?

In analytic geometry, the equation of a circle describes the set of all coordinate points $(x, y)$ that are at a fixed distance $r$ (the radius) from a fixed center point $(h, k)$. This geometric relationship stems directly from the Pythagorean theorem and the Euclidean distance formula.

Whether you are solving high school algebra problems, computer graphics algorithms, engineering blueprints, or physics orbital mechanics, circle equations allow you to determine intersections, tangents, areas, and perimeters. You can also explore our Circle Center Calculator and Circle Calculator for related circular geometry tools.

Forms of the Circle Equation

1. Standard Form (Center-Radius Form)

The most common and intuitive form of the circle equation is:

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

Where:

$(h, k)$ are the coordinates of the circle center
$r$ is the circle radius ($r > 0$)
If the circle center is at the origin $(0, 0)$, the equation simplifies to $x^2 + y^2 = r^2$

2. General Form

Expanding the standard form and collecting like terms produces the general quadratic form of a circle:

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

Or with an arbitrary leading coefficient $A$:

$$Ax^2 + Ay^2 + Bx + Cy + D = 0 \quad (A \neq 0)$$

From the general form, the center and radius are retrieved by completing the square:

$$h = -\frac{D}{2}, \quad k = -\frac{E}{2}, \quad r = \sqrt{h^2 + k^2 - F}$$

3. Parametric Form

Expressing the coordinate position $(x, y)$ as a function of the parameter $t \in [0, 2\pi]$ (the angle in radians):

$$\begin{cases} x(t) = h + r\cos(t) \\ y(t) = k + r\sin(t) \end{cases}$$

How to Find X and Y Intercepts of a Circle

The intercepts of a circle are the points where the circle crosses the coordinate axes:

X-intercepts (set $y = 0$): $$(x - h)^2 + (-k)^2 = r^2 \implies (x - h)^2 = r^2 - k^2 \implies x = h \pm \sqrt{r^2 - k^2}$$ If $r^2 > k^2$, there are two $x$-intercepts. If $r^2 = k^2$, the circle is tangent to the $x$-axis at $(h, 0)$. If $r^2 < k^2$, there are no real $x$-intercepts.
Y-intercepts (set $x = 0$): $$(-h)^2 + (y - k)^2 = r^2 \implies (y - k)^2 = r^2 - h^2 \implies y = k \pm \sqrt{r^2 - h^2}$$ If $r^2 > h^2$, there are two $y$-intercepts. If $r^2 = h^2$, the circle is tangent to the $y$-axis at $(0, k)$. If $r^2 < h^2$, there are no real $y$-intercepts.

Step-by-Step Example: Converting General Form to Standard Form

Convert the equation $x^2 + y^2 - 4x + 6y - 12 = 0$ to standard form and find its center, radius, and intercepts.

Group $x$ and $y$ terms and move the constant to the right: $$(x^2 - 4x) + (y^2 + 6y) = 12$$
Complete the square for $x$ and $y$: Take half of the $x$-coefficient $(-4/2 = -2)$, square it $((-2)^2 = 4)$. Take half of the $y$-coefficient $(6/2 = 3)$, square it $(3^2 = 9)$. Add $4 + 9 = 13$ to both sides: $$(x^2 - 4x + 4) + (y^2 + 6y + 9) = 12 + 4 + 9$$
Factor into binomial squares: $$(x - 2)^2 + (y + 3)^2 = 25$$
Identify center and radius: Center $(h, k) = (2, -3)$, Radius $r = \sqrt{25} = 5$.
Find $x$-intercepts (set $y = 0$): $$(x - 2)^2 + (0 + 3)^2 = 25 \implies (x - 2)^2 + 9 = 25 \implies (x - 2)^2 = 16 \implies x = 2 \pm 4$$ The $x$-intercepts are $(6, 0)$ and $(-2, 0)$.
Find $y$-intercepts (set $x = 0$): $$(0 - 2)^2 + (y + 3)^2 = 25 \implies 4 + (y + 3)^2 = 25 \implies (y + 3)^2 = 21 \implies y = -3 \pm \sqrt{21}$$ The $y$-intercepts are $(0, -3 + \sqrt{21}) \approx (0, 1.58)$ and $(0, -3 - \sqrt{21}) \approx (0, -7.58)$.

Frequently Asked Questions

What is the standard equation of a circle?

The standard form equation of a circle is $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the center of the circle and $r$ is the radius.

How do you find the equation of a circle given two diameter endpoints?

First, find the center $(h, k)$ by calculating the midpoint of the two endpoints: $h = (x_1 + x_2)/2$ and $k = (y_1 + y_2)/2$. Second, compute the radius $r = \frac{1}{2}\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. Finally, plug $(h, k)$ and $r$ into $(x - h)^2 + (y - k)^2 = r^2$.

What does a circle equation with radius 0 represent?

An equation of the form $(x - h)^2 + (y - k)^2 = 0$ represents a degenerate circle, which is simply a single point $(h, k)$.

How do you tell if a general quadratic equation is a circle?

An equation $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$ represents a circle if and only if: (1) $A = C \neq 0$ (coefficients of $x^2$ and $y^2$ are equal and non-zero), (2) $B = 0$ (no $xy$ cross-term), and (3) $D^2 + E^2 - 4AF > 0$ (radius is real and positive).

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