Equation of a Circle Calculator

Enter center and radius, three points on the circle, or general form coefficients to find the equation of a circle in both standard and general form.

⭕ Equation of a Circle Calculator
Center x-coordinate (h)0
−5050
Center y-coordinate (k)0
−5050
Radius (r)5
units
0.150
Point 1
x y
Point 2
x y
Point 3
x y

x² + y² + Dx + Ey + F = 0

Coefficient D (x term)
Coefficient E (y term)
Constant F
Standard Form
General Form
Center (h, k)
Radius
Diameter
Area
Circumference

⭕ What is the Equation of a Circle?

The equation of a circle is a mathematical expression that describes all points in a plane that are a fixed distance (the radius) from a fixed point (the center). The standard form is (x − h)² + (y − k)² = r², where (h, k) is the center and r is the radius. This equation is derived directly from the Pythagorean theorem applied to the distance from any point (x, y) on the circle to the center (h, k).

There are three common situations where you need to find the equation of a circle. First, you may know the center and radius directly, in which case you substitute into the standard form immediately. Second, you may have three specific points that lie on the circle (for example, three measured coordinates from a physical object or a geometry problem), in which case you solve a system of linear equations to recover h, k, and r. Third, you may have a circle equation given in expanded general form x² + y² + Dx + Ey + F = 0, and you need to convert it back to standard form by completing the square.

The equation of a circle is widely used across mathematics and applied fields. In analytic geometry, it is the simplest conic section. In physics, circular motion and orbital paths are described using circle equations in Cartesian or polar coordinates. In computer graphics, circles and arcs are defined by their equations for rasterization and collision detection. In GPS and telecommunications, signal coverage areas are modeled as circles in 2D coordinate systems, and the circle equation determines whether a point (a receiver) is within range.

A key distinction is between the circle (the boundary, the set of points at exactly distance r from center) and the disk (the filled region, all points at distance at most r). The equation of a circle describes the boundary. The general form x² + y² + Dx + Ey + F = 0 can also describe a single point (if r² = 0) or have no real solutions (if r² is negative, called an imaginary circle). This calculator checks for these cases and reports an error if the coefficients do not describe a real circle.

📐 Formula

(x − h)² + (y − k)² = r²
(h, k) = center of the circle
r = radius of the circle
Example: Center (3, −2), radius 5: (x − 3)² + (y + 2)² = 25
x² + y² + Dx + Ey + F = 0
h = −D/2,   k = −E/2
r² = D²/4 + E²/4 − F  (must be positive for a real circle)
Three-point method: Subtract pairs of circle equations to get two linear equations, then solve for h and k by Cramer's rule.

📖 How to Use This Calculator

Steps

1
Choose a mode - Click Center and Radius if you know h, k, and r. Click Three Points if you have three coordinates on the circle. Click General Form if you have coefficients D, E, F from the expanded equation.
2
Enter the inputs - In Center and Radius mode, use the sliders or type values. In Three Points mode, enter x and y for all three points. In General Form mode, enter D, E, and F (all can be negative or zero).
3
Click Calculate - The calculator shows the standard form equation, the expanded general form, center coordinates, radius, diameter, area, and circumference.
4
Use the equations - Copy the standard or general form for use in homework, reports, or code. Use the permalink button to share this specific circle with others.

💡 Example Calculations

Example 1 - Circle with center (3, −2) and radius 5

Find the standard and general form for center (3, −2), radius 5

1
Standard form: substitute h = 3, k = −2, r = 5. Result: (x − 3)² + (y + 2)² = 25.
2
Expand to general form: x² − 6x + 9 + y² + 4y + 4 = 25. Rearrange: x² + y² − 6x + 4y − 12 = 0. So D = −6, E = 4, F = −12.
3
Area = pi × 25 = 78.5398 sq units. Circumference = 2 × pi × 5 = 31.4159 units.
Result: Standard: (x − 3)² + (y + 2)² = 25, General: x² + y² − 6x + 4y − 12 = 0
Try this example →

Example 2 - Circle through three points (1, 0), (−1, 0), (0, 1)

Find the circle through (1, 0), (−1, 0), and (0, 1)

1
Subtract equations for points 1 and 2: 2(x2 − x1)h + 2(y2 − y1)k = x2² − x1² + y2² − y1². For (1,0) and (−1,0): −4h + 0k = 0, so h = 0.
2
Subtract equations for points 2 and 3: for (−1,0) and (0,1): 2h + 2k = 1. With h = 0: k = 0.5.
3
Radius from point (1, 0) to center (0, 0.5): r = sqrt(1 + 0.25) = sqrt(1.25) = 1.118 units. Equation: x² + (y − 0.5)² = 1.25.
Result: Center (0, 0.5), radius 1.118, equation: x² + (y − 0.5)² = 1.25
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Example 3 - Convert general form x² + y² − 4x + 6y − 3 = 0

Find center and radius from D = −4, E = 6, F = −3

1
Center: h = −D/2 = −(−4)/2 = 2. k = −E/2 = −6/2 = −3.
2
Radius squared: r² = D²/4 + E²/4 − F = 4 + 9 − (−3) = 4 + 9 + 3 = 16. Radius r = 4.
3
Standard form: (x − 2)² + (y + 3)² = 16. Area = pi × 16 = 50.27 sq units. Circumference = 2 × pi × 4 = 25.13 units.
Result: Center (2, −3), radius 4, standard form: (x − 2)² + (y + 3)² = 16
Try this example →

❓ Frequently Asked Questions

What is the standard form of the equation of a circle?+
The standard form is (x - h) squared + (y - k) squared = r squared, where (h, k) is the center and r is the radius. This form directly reveals the center and radius at a glance. For example, (x - 5) squared + (y + 2) squared = 49 has center (5, -2) and radius 7. The sign of h is opposite in the equation: (x - h) means the center is at h, not -h.
How do I find the equation of a circle with diameter endpoints?+
Compute the center as the midpoint of the two endpoints: h = (x1 + x2) / 2, k = (y1 + y2) / 2. Compute the radius as half the distance between them: r = sqrt((x2 - x1) squared + (y2 - y1) squared) / 2. Then enter h, k, r into Center and Radius mode. Example: endpoints (1, 3) and (7, 11): center (4, 7), radius = sqrt(36 + 64) / 2 = 10 / 2 = 5.
What is the equation of the unit circle?+
The unit circle has center (0, 0) and radius 1. Its equation is x squared + y squared = 1. Every point on the unit circle has coordinates (cos theta, sin theta) for some angle theta. The unit circle is the foundation of trigonometry: sine and cosine are defined as the y and x coordinates of the point on the unit circle at angle theta from the positive x-axis.
How do you convert general form to standard form?+
Complete the square. Given x squared + y squared + Dx + Ey + F = 0, group and complete: (x + D/2) squared + (y + E/2) squared = D squared / 4 + E squared / 4 - F. The center is (-D/2, -E/2) and r squared = D squared / 4 + E squared / 4 - F. If r squared is negative, no real circle exists. If r squared = 0, it is a single point.
How many points are needed to uniquely define a circle?+
Three non-collinear points uniquely determine a circle. One point gives infinitely many circles (any circle passing through that point). Two distinct points still leave infinitely many circles, all with centers on the perpendicular bisector of the segment. The third non-collinear point narrows it to exactly one circle. This is why the Three Points mode requires all three inputs to be distinct and non-collinear.
What does the general form x squared + y squared + Dx + Ey + F = 0 tell you?+
The general form is the expanded version of the standard form. The coefficients D, E, F encode the center and radius: h = -D/2, k = -E/2, and r squared = D squared / 4 + E squared / 4 - F. Note that in the general form, the coefficients of x squared and y squared must both be 1 (or the equation must be divided through). If r squared computed from D, E, F is negative, the equation has no real solutions.
Can a circle equation have negative r squared?+
Yes. The general form x squared + y squared + Dx + Ey + F = 0 can have r squared = D squared / 4 + E squared / 4 - F negative. In that case, no real point (x, y) satisfies the equation. This is called an imaginary circle. It arises mathematically but has no physical meaning. This calculator detects this case and reports an error: These coefficients do not describe a real circle (r squared must be positive).
How is a circle equation used in computer science?+
In computer graphics, the circle equation is used for rasterization (Bresenham's circle algorithm), clipping, and collision detection in games. A point P is inside a circle if (Px - h) squared + (Py - k) squared is less than r squared. Collision detection between a circle and a point runs in O(1) time using this test. The equation is also used in GPS coverage modeling, where a transmitter covers a circular area around its location.
What is the equation of a circle tangent to the x-axis?+
A circle is tangent to the x-axis when the distance from the center to the x-axis equals the radius. If the center is (h, k), tangency to the x-axis means the absolute value of k equals r. So the equation is (x - h) squared + (y - k) squared = k squared. For a circle with center (3, 4) tangent to the x-axis: r = 4, equation is (x - 3) squared + (y - 4) squared = 16.
What is the polar equation of a circle?+
A circle centered at the origin with radius r has polar equation rho = r (a constant), where rho is the distance from the origin. A circle centered at (a, 0) with radius a passing through the origin has polar equation rho = 2a cos(theta). A circle centered at (0, b) passing through the origin has rho = 2b sin(theta). The polar form is useful in physics for circular orbit problems.
How is a circle different from an ellipse in equation form?+
A circle is a special ellipse where both semi-axes are equal. The standard ellipse equation is (x - h) squared / a squared + (y - k) squared / b squared = 1. When a = b = r, this becomes (x - h) squared / r squared + (y - k) squared / r squared = 1, which multiplies out to (x - h) squared + (y - k) squared = r squared, the standard circle equation. An ellipse with unequal a and b is stretched along one axis and compressed along the other.