Audited 05 Aug 2026·Last updated 28 Oct 2026·3 citations·Tier 2·0 uses

Standard Equation Circle Calculator

Write a circle as (x − h)² + (y − k)² = r² from its centre and radius, and get r² plus the constant F of the expanded x² + y² + Dx + Ey + F = 0 form.

Standard Equation Circle Calculator

Right side r squared
25
The right-hand side of the standard form, the radius squared, in the square of your coordinate unit.
Equivalent general-form F
-12

Background.

Start with the simplest circle there is, x² + y² = r², centred on the origin. To slide it so its centre sits at (h, k), replace x with x − h and y with y − k. The result, (x − h)² + (y − k)² = r², is the standard form, and its minus signs are simply the cost of the shift: a centre two units right and three down gives (x − 2)² and (y + 3)², which looks backwards until you notice that x − 2 measures how far a point lies to the right of the centre.

In this form the circle can be read without any algebra. The centre sits in the brackets, the radius is the square root of the right-hand side, and the four extreme points, left, right, top and bottom, are one step away at (h ± r, k) and (h, k ± r). Where the circle crosses an axis comes out with a square root rather than the quadratic formula, because setting y = 0 or x = 0 leaves a single squared bracket to undo.

Give the page the centre coordinates h and k and a positive radius, and it returns the right-hand side, r², together with the constant F of the same circle multiplied out as x² + y² + Dx + Ey + F = 0, where F = h² + k² − r². The other two expanded coefficients need no calculation, D = −2h and E = −2k, so they are left for you to write in.

The radius must be greater than zero. A zero radius describes a single point, not a circle, and a negative one has no geometric meaning, so both are refused. Fitting a circle to measured points, or finding one through given points, is a different job from writing out a circle you already know.

What is standard equation circle calculator?

The standard equation of a circle, (x − h)² + (y − k)² = r², is the test a point (x, y) must pass to sit exactly r units from the centre (h, k). The left side is the square of that point's distance from the centre, built from its horizontal and vertical offsets; the right side is the square of the radius. Where the left side comes out smaller the point is inside the circle, and where it comes out larger the point is outside.

How to use this calculator.

  1. Enter the centre's x-coordinate as h and its y-coordinate as k, each with its own sign. A centre at (2, −3) goes in as 2 and −3; the sign flip that appears inside the brackets is something you write, not something you enter.
  2. Enter the radius as a positive number. If you were given a diameter, halve it first; if you were given r² from an existing equation, take its square root.
  3. Write the standard form from h, k and the r² shown. For the expanded form, use D = −2h, E = −2k and the F returned here in x² + y² + Dx + Ey + F = 0.
  4. Check by substituting a point you know is on the circle, such as (h + r, k), into either form; both should balance exactly.

The formula.

(x-h)^2+(y-k)^2=r^2; expanded form x^2+y^2-2hx-2ky+h^2+k^2-r^2=0

The standard form is (x − h)² + (y − k)² = r², so the right-hand side is just the radius squared. Multiplying out the brackets gives x² − 2hx + h² + y² − 2ky + k² = r², and gathering everything on one side gives x² + y² − 2hx − 2ky + (h² + k² − r²) = 0. The constant term F = h² + k² − r² is the second number the page returns. Because F folds the centre and radius together, it grows quickly as the centre moves away from the origin: a radius-5 circle centred at (100, 0) has F = 9,975, while the same circle at (2, −3) has F = −12. F is the squared distance from the origin to the centre minus r², so its sign places the origin: inside the circle when F is negative, on it when F is zero, outside when F is positive.

A worked example.

Example

The example circle has centre (2, −3) and radius 5. The right-hand side is r² = 5² = 25, so the standard form is (x − 2)² + (y + 3)² = 25, with the −3 turning into + 3 inside the bracket. For the expanded form, F = 2² + (−3)² − 5² = 4 + 9 − 25 = −12, and with D = −2 × 2 = −4 and E = −2 × (−3) = 6 the circle becomes x² + y² − 4x + 6y − 12 = 0. F is negative, so the origin lies inside: it is √13, about 3.606, from the centre, less than the radius of 5. Setting y = 0 in the standard form leaves (x − 2)² = 16, so the circle cuts the x-axis at x = 6 and x = −2. Shrink the radius to 3 and F becomes 4 + 9 − 9 = 4, positive, because the origin now falls outside the circle.

center Y-3
center X2
radius5

Frequently asked questions.

Why does the right-hand side say 25 when my radius is 5?
The standard form compares squared distances, so the right-hand side is r², not r. Writing (x − 2)² + (y + 3)² = 5 is a common slip, and it describes a much smaller circle with radius √5, about 2.236. When reading an equation, always take the square root of the right side to get the radius.
How do I graph the circle on a calculator that only plots y = …?
Solve for y to get two halves: y = k + √(r² − (x − h)²) for the top and y = k − √(r² − (x − h)²) for the bottom. For the example, enter y = −3 + √(25 − (x − 2)²) and y = −3 − √(25 − (x − 2)²). Use a square viewing window, or the circle will look like an ellipse.
How do I find where the circle crosses the axes?
Set the other coordinate to zero and undo the square. For the example, y = 0 gives (x − 2)² = 25 − 9 = 16, so x = 6 or −2. x = 0 gives (y + 3)² = 21, so y is about 1.583 or −7.583. If the number left under the square root is negative, the circle never reaches that axis.
Is there a parametric version of the same circle?
Yes: x = h + r cos t and y = k + r sin t, with t running from 0 to 2π radians. It traces the circle once, anticlockwise from the rightmost point, and is the easiest form to use when plotting points in a spreadsheet or program. For the example, t = 0 gives (7, −3) and t = π/2 gives (2, 2).
Why won't the page accept a radius of zero?
With r = 0 the equation (x − h)² + (y − k)² = 0 is satisfied only by the centre itself, a single point sometimes called a point circle. A negative r² would have no real points at all. The page requires a positive radius so that the equation it writes out is a genuine circle.

How this page was produced

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Method
(x-h)^2+(y-k)^2=r^2; expanded form x^2+y^2-2hx-2ky+h^2+k^2-r^2=0
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