How do you find the equation of a circle from two points?

Two points alone do not determine a unique circle — infinitely many circles pass through any two points. You need a third point on the circle, or additional information such as the center or the radius. With three non-collinear points, substitute each into $(x - h)^2 + (y - k)^2 = r^2$ to create a system of three equations in three unknowns ($h$, $k$, and $r^2$), then solve.

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

Standard form is $(x - h)^2 + (y - k)^2 = r^2$; you can read the center and radius directly. General form is $x^2 + y^2 + Dx + Ey + F = 0$, which results from expanding and rearranging the standard form. To convert general form back to standard form, complete the square on $x$ and $y$.

How do you tell if a point is inside, on, or outside a circle?

Plug the point's coordinates into the left side of the standard-form equation. If the result equals $r^2$, the point is on the circle. If it is less than $r^2$, the point is inside. If it is greater than $r^2$, the point is outside.

Equation of a Circle — Definition, Formula & Examples

The equation of a circle is an algebraic equation that describes all the points at a fixed distance (the radius) from a central point (the center) on a coordinate plane. In standard form, it is written as

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

, where

(h, k)

is the center and

r

is the radius.

A circle is the locus of all points $(x, y)$ in the Cartesian plane that are equidistant from a fixed point called the center. If the center is $(h, k)$ and the constant distance is $r > 0$ , the standard equation is $(x - h)^2 + (y - k)^2 = r^2$ . This equation is derived directly from the distance formula. An equivalent representation, called the general form, is $x^2 + y^2 + Dx + Ey + F = 0$ , obtained by expanding and rearranging the standard form.

Key Formula

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

Where:

$x$ = the x-coordinate of any point on the circle
$y$ = the y-coordinate of any point on the circle
$h$ = the x-coordinate of the center
$k$ = the y-coordinate of the center
$r$ = the radius of the circle (must be positive)

How It Works

Start with the standard form

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

. The values

h

and

k

tell you where the center sits on the coordinate plane, and

r

tells you how far every point on the circle is from that center. To write the equation, you need two pieces of information: the center and the radius (or enough data to find them). If you are given the general form

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

, you can convert it to standard form by completing the square on both

x

and

y

. A point

(a, b)

lies on the circle if and only if substituting

x = a

and

y = b

into the equation produces a true statement. When the center is at the origin, the equation simplifies to

x^2 + y^2 = r^2

Worked Example

Problem: Write the equation of a circle with center (3, −2) and radius 5.

Step 1: Identify the center and radius. Here

h = 3

k = -2

, and

r = 5

Step 2: Substitute these values into the standard form

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

(x - 3)^2 + (y - (-2))^2 = 5^2

Step 3: Simplify. Subtracting a negative becomes addition, and

5^2 = 25

(x - 3)^2 + (y + 2)^2 = 25

Answer:

(x - 3)^2 + (y + 2)^2 = 25

Another Example

This example starts from the general form and requires completing the square — a key skill when the equation is not already in standard form.

Problem: Find the center and radius of the circle given by

x^2 + y^2 - 6x + 4y - 12 = 0

Step 1: Group the

x

terms and

y

terms together, and move the constant to the right side.

(x^2 - 6x) + (y^2 + 4y) = 12

Step 2: Complete the square for

x

. Take half of

-6

, which is

-3

, and square it to get

9

. Add

9

to both sides.

(x^2 - 6x + 9) + (y^2 + 4y) = 12 + 9

Step 3: Complete the square for

y

. Take half of

4

, which is

2

, and square it to get

4

. Add

4

to both sides.

(x^2 - 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4

Step 4: Factor each perfect-square trinomial and simplify the right side.

(x - 3)^2 + (y + 2)^2 = 25

Step 5: Read off the center and radius. The center is

(3, -2)

and

r^2 = 25

, so

r = 5

Answer: Center

(3, -2)

, radius

5

Why It Matters

The equation of a circle appears throughout Geometry, Algebra II, and Precalculus courses, and it is a staple on the SAT and ACT. Engineers and physicists use circle equations when modeling orbits, designing gears, and computing signal ranges. It also serves as the gateway to studying all conic sections — ellipses, parabolas, and hyperbolas.

Common Mistakes

Mistake: Mixing up the signs of

h

and

k

. For example, seeing

(x + 3)^2

and writing the center's

x

-coordinate as

+3

Correction: The standard form uses subtraction:

(x - h)^2

. If you see

(x + 3)^2

, rewrite it as

(x - (-3))^2

, so

h = -3

Mistake: Using

r

instead of

r^2

on the right side of the equation.

Correction: The standard form has

r^2

on the right. If the radius is 5, the right side should be 25, not 5.

Mistake: Forgetting to add the completing-the-square constant to both sides.

Correction: When you add a value inside one set of parentheses on the left, you must add the same value to the right side to keep the equation balanced.

Check Your Understanding

Write the standard-form equation of a circle centered at

(-1, 4)

with radius

3

Hint: Remember that

(x - (-1))

simplifies to

(x + 1)

Answer:

(x + 1)^2 + (y - 4)^2 = 9

Find the center and radius:

x^2 + y^2 + 10x - 2y + 17 = 0

Hint: Half of 10 is 5 (square it to get 25); half of

-2

-1

(square it to get 1).

Answer: Center

(-5, 1)

, radius

3

. After completing the square:

(x+5)^2 + (y-1)^2 = 9

Is the point

(7, -2)

inside, on, or outside the circle

(x - 3)^2 + (y + 2)^2 = 25

Hint: Compute the left side and compare to 25.

Answer: On the circle, because

(7-3)^2 + (-2+2)^2 = 16 + 0 = 16

. Wait —

16 < 25

, so the point is actually inside the circle.

Related Terms

Conic Sections — Parent family that includes circles
Ellipse — Generalization of a circle with two axes
Hyperbola — Another conic section, uses subtraction
Focus (conic section) — Key point for all conics; circle has coinciding foci
Degenerate — Circle with radius 0 is a degenerate case
Double Cone — Surface from which conic sections are sliced
Circle Graph (Pie Chart) — Uses circles in data visualization
Major Axis of an Ellipse — Becomes the diameter when the ellipse is a circle