Equation of a Circle: Formula, Center & Radius Explained

October 13, 2026

The equation of a circle names every point in the plane that sits at one fixed distance from a fixed center. If the center is $(h, k)$ and the radius is $r > 0$ , a point $(x, y)$ lies on the circle exactly when its distance to the center is $r$ .

The live board below freezes a center $C$ and a radius slider so you can watch $r$ move the circle, not only memorize a number in a formula.

Slider r sets the radius. Center C=(1,1); P sits at (1+r, 1) on the circle. Move the slider — the circle grows and shrinks, and every point still satisfies (x−1)²+(y−1)²=r².

From distance to standard form

Distance from $(x, y)$ to $(h, k)$ is

$(x - h)^{2} + (y - k)^{2} .$

Set that equal to $r$ and square both sides (both sides are positive). The same set of points satisfies

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

That is the standard form of the equation of a circle. It is the radius condition written so a Cartesian plane can check it. The handbook builds the same freeze: one center and one length, then the squared Cartesian equation.

When the center sits at the origin and $r = 1$ , the equation collapses to $x^{2} + y^{2} = 1$ — the unit circle. Every other circle in the plane is a translate and a scale of that one.

How to find the equation of a circle

You need the center and the radius. Then plug them into standard form.

Given center and radius. Center $(h, k)$ , radius $r$ → write $(x - h)^{2} + (y - k)^{2} = r^{2}$ .
Given center and a point on the circle. The radius is the distance from the center to that point; then use step 1.
Given an expanded (general) form. The equation arrives as $x^{2} + y^{2} + D x + E y + F = 0$ . Complete the square to recover $h$ , $k$ , and $r$ . Example 2 does that arithmetic.

Worked examples

Example 1 — write the equation. Center $C = (1, 1)$ , radius $r = 3$ (the opening position on the board: $P$ sits at $(4, 1)$ ).

$(x - 1)^{2} + (y - 1)^{2} = 9.$

Check: the point $(4, 1)$ gives $(4 - 1)^{2} + (1 - 1)^{2} = 9$ , so it lies on the circle. Any other point at distance $3$ from $(1, 1)$ works the same way.

Example 2 — read center and radius. Start from

$(x + 2)^{2} + (y - 3)^{2} = 16.$

Here $h = - 2$ , $k = 3$ , and $r = 4$ . The center is $(- 2, 3)$ ; the radius is $4$ . Expanding puts the same circle in general form:

$x^{2} + y^{2} + 4 x - 6 y - 3 = 0.$

Complete the square to come back. Move the constant, then add the squares of half the linear coefficients ( $4/2 = 2$ , so add $4$ ; $- 6/2 = - 3$ , so add $9$ ):

$x^{2} + 4 x + y^{2} - 6 y = 3,$

$(x + 2)^{2} + (y - 3)^{2} = 3 + 4 + 9 = 16.$

The center is $(- 2, 3)$ and the radius is $4$ again.

Example 3 — from two pieces of data. Center at $(0, 0)$ and a point $(3, 4)$ on the circle. Then $r = 3^{2} + 4^{2} = 5$ , so

$x^{2} + y^{2} = 25.$

Move the slider, keep the equation

$C$ is fixed at $(1, 1)$ . The slider $r$ is the radius. Point $P$ sits at $(1 + r, 1)$ on the circle, so the segment $C P$ is always length $r$ . Move the slider. The circle grows or shrinks with $r$ , and every point on the curve still satisfies $(x - 1)^{2} + (y - 1)^{2} = r^{2}$ . The center stays $(1, 1)$ ; only the squared radius in the equation changes.

That is the check that the figure matches the formula: one center, one length, and every point on the curve is at that length from the center. On the handbook page you can drag the center, with the tangent perpendicular to the radius. Area and circumference for related shapes live with circles, ellipses, and sectors. Distance in coordinates sits with the broader plane analytic geometry cluster when you need the two-point formula before squaring.

GoodMaith keeps a relation like this as an object on a live board: name a center and a radius, move a slider, and see why $(x - h)^{2} + (y - k)^{2} = r^{2}$ is exactly the list of points at distance $r$ from $(h, k)$ . The answer was never only the words “equation of a circle”; it was seeing one radius force every point on the curve.