How do I find the equation of a circle if I know the center and two points on the circle?

Use the Distance Formula to calculate the distance from the center to either of the two points; this distance is the radius r. Then substitute the center (h, k) and the computed radius into (x - h)^{2} + (y - k)^{2} = r^{2}.

What is the difference between the circle equation (x - h)^{2} + (y - k)^{2} = r^{2} and the inequality (x - h)^{2} + (y - k)^{2} < r^{2}?

The equation represents points exactly on the circle's boundary. The inequality (x - h)^{2} + (y - k)^{2} < r^{2} represents all points strictly inside the circle, and \le r^{2} includes both the interior and the boundary.

Does the circle equation have a special form when the center is at the origin?

Yes, when (h, k) = (0, 0), the equation simplifies to x^{2} + y^{2} = r^{2}. This is the most basic form and appears frequently in introductory geometry and trigonometry.

What does it mean if the radius squared is negative, like r^{2} = -4?

It means the circle does not exist as a real geometric object; the equation has no solution. If r^{2} = 0, the circle degenerates into a single point at the center (h, k).

Equation of a Circle

Q: What is the difference between the circle equation (x - h)^{2} + (y - k)^{2} = r^{2} and the inequality (x - h)^{2} + (y - k)^{2} < r^{2}?

The equation represents points exactly on the circle's boundary. The inequality (x - h)^{2} + (y - k)^{2} < r^{2} represents all points strictly inside the circle, and \le r^{2} includes both the interior and the boundary.

Q: Does the circle equation have a special form when the center is at the origin?

Yes, when (h, k) = (0, 0), the equation simplifies to x^{2} + y^{2} = r^{2}. This is the most basic form and appears frequently in introductory geometry and trigonometry.

Q: What does it mean if the radius squared is negative, like r^{2} = -4?

It means the circle does not exist as a real geometric object; the equation has no solution. If r^{2} = 0, the circle degenerates into a single point at the center (h, k).

Describes a circle using its center and radius; essential for finding circle equations, checking if points lie on circles, or solving circle geometry problems.

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

Solve a problem with Equation of a Circle

Type the problem. The solver will use Equation of a Circle where Equation of a Circle is the right tool, and tell you when it is not.

What each symbol means

What Equation of a Circle takes

Symbol	Meaning
$h$	The x-coordinate of the circle's center; confusing it with the radius or y-coordinate will misplace the circle horizontally.
$k$	The y-coordinate of the circle's center; swapping it with h flips the circle's displacement along the y-axis instead of the x-axis.
$r$	The radius, the distance from the circle's center to any point on the circle, always positive; forgetting to square it produces a degenerate shape instead of a full circle.
$x$	The x-coordinate of any point that lies on the circle; it is a variable representing infinitely many possible values, not a single fixed number.
$y$	The y-coordinate of any point on the circle; it pairs with x to specify which points satisfy the circle's equation, and swapping them reflects the circle across the line $y = x$ .

When to use it

When you need to write a circle's equation from its center and radius, or determine whether a point lies on, inside, or outside a specific circle.

Level

Usually taught in: Geometry · Appears on: SAT

Worked examples

1. Write the equation of a circle given center and radius

Problem

Find the equation of a circle with center at

(3, 2)

and radius

5

$\text{Given:} \quad (h, k) = (3, 2), \quad r = 5$
We identify the center coordinates and radius from the problem statement.
$(x - h)^{2} + (y - k)^{2} = r^{2}$
We write the standard form of the circle equation before substituting values.
$(x - 3)^{2} + (y - 2)^{2} = 5^{2}$
We substitute $h = 3$ , $k = 2$ , and $r = 5$ directly into the formula.
$(x - 3)^{2} + (y - 2)^{2} = 25$
We compute $5^{2} = 25$ to simplify the right side.

Answer: $(x - 3)^{2} + (y - 2)^{2} = 25$

This is a straightforward substitution using the standard form. Any point $(x, y)$ satisfying this equation is exactly 5 units away from the center $(3, 2)$ .

2. Handle negative center coordinates and a square-root radius

Problem

Write the equation of a circle centered at

(-2, 3)

with radius

\sqrt{13}

$\text{Given:} \quad (h, k) = (-2, 3), \quad r = \sqrt{13}$
We note that $h = -2$ is negative; this will affect how we write the $(x - h)$ term.
$(x - h)^{2} + (y - k)^{2} = r^{2}$
We use the standard form, remembering that subtraction will handle the negative center coordinate.
$(x - (-2))^{2} + (y - 3)^{2} = (\sqrt{13})^{2}$
We substitute $h = -2$ , $k = 3$ , and $r = \sqrt{13}$ into the formula.
$(x + 2)^{2} + (y - 3)^{2} = 13$
We simplify: $x - (-2) = x + 2$ (double negative becomes addition) and $(\sqrt{13})^{2} = 13$ .

Answer: $(x + 2)^{2} + (y - 3)^{2} = 13$

This example reinforces that $(x - h)$ automatically handles negative center coordinates through subtraction—you do not flip signs. Also note that the radius can be irrational, but squaring it often yields a rational or integer value.

3. Model a physical boundary: heat-dissipation region from a lab heat source

Problem

A scientist studying thermal diffusion places a heat source at coordinates

(4, 5)

on a lab table (in meters). She measures that the temperature drops to half its peak value at all points exactly 3 meters away from the source. Write the equation describing this half-temperature boundary.

$\text{Center (heat source):} (h, k) = (4, 5), \quad \text{Boundary radius:} r = 3 \text{ m}$
The half-temperature boundary forms a circle centered at the heat source with radius 3 meters.
$(x - h)^{2} + (y - k)^{2} = r^{2}$
We apply the circle equation, which models all points at a fixed distance from the source.
$(x - 4)^{2} + (y - 5)^{2} = 3^{2}$
We substitute the center coordinates $(4, 5)$ and radius $3$ into the standard form.
$(x - 4)^{2} + (y - 5)^{2} = 9$
We compute $3^{2} = 9$ to complete the equation.

Answer: $(x - 4)^{2} + (y - 5)^{2} = 9$

This real-world application shows how the circle equation models physical phenomena. Any point $(x, y)$ on this circle is exactly 3 meters from the heat source and marks where the temperature reaches half its peak. Points inside the circle are warmer; points outside are cooler.

Common mistakes

Where Equation of a Circle usually goes wrong

The mistake	Why it is wrong	The fix
Writing $(x - h)^{2} + (y - k)^{2} = r$ instead of $= r^{2}$ .	Omitting the square of the radius completely changes what the equation represents and destroys its geometric meaning as a circle.	Always square the radius: $(x - h)^{2} + (y - k)^{2} = r^{2}$ . The $r^{2}$ on the right side is essential.
Writing $(x + h)^{2} + (y + k)^{2} = r^{2}$ when the center is $(-h, -k)$ .	The subtraction in $(x - h)$ automatically accounts for negative values; using addition with a negative $h$ reverses the sign again and places the circle entirely in the wrong location.	Always write $(x - h)$ and $(y - k)$ regardless of the signs of $h$ and $k$ . Subtraction handles both positive and negative center coordinates correctly.
Expanding the equation to $x^{2} + y^{2} - 2hx - 2ky + h^{2} + k^{2} = r^{2}$ and losing track of the center when reading the equation backward.	The expanded form obscures the geometric meaning—the center $(h, k)$ and radius $r$ are no longer visible at a glance, making it easy to misidentify them later.	Keep the equation in standard form $(x - h)^{2} + (y - k)^{2} = r^{2}$ as your primary representation so the center and radius remain explicit and easy to read.

Tips and when to use something else

Substitute a known point into the equation to verify it lies on the circle: if the left side equals $r^{2}$ , you are correct; if not, recheck your center or radius.
To check if a point is inside or outside a circle without graphing, substitute it into $(x - h)^{2} + (y - k)^{2}$ : if the result is less than $r^{2}$ , the point is inside; if greater, it is outside.
When the center is at the origin, the equation simplifies to $x^{2} + y^{2} = r^{2}$ —this is a special case that appears frequently, so memorize it as a shortcut.
If you encounter a circle equation that is expanded or rearranged, use completing the square on both $x$ and $y$ terms to recover the standard form and identify the center and radius.

Related formulas

Distance FormulaCalculates the shortest distance between two points on a coordinate plane using their coordinates and the Pythagorean theorem.Midpoint FormulaThe midpoint formula finds the coordinates of the point halfway between two endpoints on a line, given their x and y coordinates.Translating a GraphTranslate graphs using y = f(x - h) + k to shift any function horizontally by h and vertically by k units, needed when repositioning known curves.Equation of an EllipseEquation of an Ellipse describes an oval-shaped curve given its center, horizontal stretch, and vertical stretch; use it to model or graph any elliptical shape.Graphing Linear InequalitiesGraphing linear inequalities shows which coordinate points satisfy a condition like y > mx + b by shading solution regions with appropriate line styles.

Frequently asked questions

How do I find the equation of a circle if I know the center and two points on the circle?: Use the Distance Formula to calculate the distance from the center to either of the two points; this distance is the radius $r$ . Then substitute the center $(h, k)$ and the computed radius into $(x - h)^{2} + (y - k)^{2} = r^{2}$ .
What is the difference between the circle equation $(x - h)^{2} + (y - k)^{2} = r^{2}$ and the inequality $(x - h)^{2} + (y - k)^{2} < r^{2}$ ?: The equation represents points exactly on the circle's boundary. The inequality $(x - h)^{2} + (y - k)^{2} < r^{2}$ represents all points strictly inside the circle, and $\le r^{2}$ includes both the interior and the boundary.
Does the circle equation have a special form when the center is at the origin?: Yes, when $(h, k) = (0, 0)$ , the equation simplifies to $x^{2} + y^{2} = r^{2}$ . This is the most basic form and appears frequently in introductory geometry and trigonometry.
What does it mean if the radius squared is negative, like $r^{2} = -4$ ?: It means the circle does not exist as a real geometric object; the equation has no solution. If $r^{2} = 0$ , the circle degenerates into a single point at the center $(h, k)$ .

Need a different method?

The full solver is not scoped to one formula — type any problem and it will pick the method.

Open the math solver

Reviewed 2026-09-18