Math Core

On the coordinate plane, a circle becomes an equation. Once you can read and write that equation, you can find a circle's center and radius at a glance, decide whether a point is inside or outside it, and graph it quickly. The key tool is one you already know: the distance formula.

From the distance formula to an equation

A circle is the set of points at a fixed distance $r$ from a center. Put the center at $(h, k)$ and let $(x, y)$ be any point on the circle. The distance formula says

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

Square both sides to remove the square root, and you get the standard equation of a circle.

Standard form of a circle

The circle with center $(h, k)$ and radius $r$ has equation

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

A circle centered at the origin has equation $x^2 + y^2 = r^2$ .

For example, the circle below has center $(2, -1)$ and radius $3$ , so its equation is

$(x - 2)^2 + (y - (-1))^2 = 3^2, \qquad\text{or}\qquad (x - 2)^2 + (y + 1)^2 = 9.$

The circle (x − 2)² + (y + 1)² = 9 has center (2, −1) and radius 3.Open in grapher →

Reading the center and radius

The equation has minus signs built in. To read the center, take the numbers being subtracted:

$(x - 4)^2$ means $h = 4$ .
$(x + 4)^2 = (x - (-4))^2$ means $h = -4$ .

The right side is $r^2$ , so the radius is its square root.

Common mistake

Two classic mistakes: flipping the signs the wrong way, and forgetting to take the square root. The circle $(x + 3)^2 + (y - 5)^2 = 16$ has center $(-3, 5)$ , not $(3, -5)$ , and radius $4$ , not $16$ .

Worked example: Reading an equation

Find the center and radius of $(x + 1)^2 + (y - 6)^2 = 50$ .

Rewrite $x + 1$ as $x - (-1)$ . The center is $(-1, 6)$ . The radius is $\sqrt{50} = 5\sqrt{2}$ , about $7.07$ .

Writing an equation

To write the equation of a circle, you need two things: the center and the radius. Often you have to find one of them first.

Worked example: Center and a point on the circle

A circle has center $(1, 2)$ and passes through $(4, 6)$ . Write its equation.

The radius is the distance from the center to the point:

$r = \sqrt{(4 - 1)^2 + (6 - 2)^2} = \sqrt{9 + 16} = 5.$

The equation is $(x - 1)^2 + (y - 2)^2 = 25$ .

Worked example: Endpoints of a diameter

The endpoints of a diameter are $(-3, 4)$ and $(5, -2)$ . Write the equation of the circle.

The center is the midpoint of the diameter:

$\left(\dfrac{-3 + 5}{2}, \dfrac{4 + (-2)}{2}\right) = (1, 1).$

The radius is the distance from the center to an endpoint: $r = \sqrt{(5 - 1)^2 + (-2 - 1)^2} = \sqrt{16 + 9} = 5$ .

The equation is $(x - 1)^2 + (y - 1)^2 = 25$ .

Tip

You don't actually need $r$ itself, only $r^2$ . In the first example, $r^2 = 3^2 + 4^2 = 25$ goes straight into the equation, no square root needed.

Completing the square

Sometimes a circle's equation arrives multiplied out, like $x^2 + y^2 - 6x + 4y - 12 = 0$ . To find the center and radius, complete the square in $x$ and in $y$ .

Worked example: Completing the square

Find the center and radius of $x^2 + y^2 - 6x + 4y - 12 = 0$ .

Group the $x$ terms and the $y$ terms, and move the constant to the right:

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

Take half of each linear coefficient and square it: $\left(\frac{-6}{2}\right)^2 = 9$ and $\left(\frac{4}{2}\right)^2 = 4$ . Add both to each side:

\begin{aligned} (x^2 - 6x + 9) + (y^2 + 4y + 4) &= 12 + 9 + 4 \\ (x - 3)^2 + (y + 2)^2 &= 25. \end{aligned}

The center is $(3, -2)$ and the radius is $5$ .

Inside, on or outside?

To decide where a point lies, substitute it into the left side of the standard equation and compare the result with $r^2$ :

less than $r^2$ : the point is inside the circle,
equal to $r^2$ : the point is on the circle,
greater than $r^2$ : the point is outside the circle.

This works because the left side is the squared distance from the point to the center.

Practice

Practice 1

Which is the equation of the circle with center $(3, -2)$ and radius $4$ ?

Practice 2

What is the center of the circle $(x + 5)^2 + (y - 1)^2 = 49$ ? Give an ordered pair.

Enter a point like (2, -3)

Practice 3

What is the radius of the circle $(x + 5)^2 + (y - 1)^2 = 49$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 4

A circle has center $(1, 2)$ and passes through the point $(-3, -1)$ . What is its radius?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 5

The endpoints of a diameter of a circle are $(-2, 3)$ and $(6, -3)$ . What is the center of the circle? Give an ordered pair.

Enter a point like (2, -3)

Practice 6

Where is the point $(4, 5)$ relative to the circle $(x - 1)^2 + (y - 1)^2 = 25$ ?

Practice 7

What is the radius of the circle $x^2 + y^2 - 6x + 8y = 0$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 8

What is the center of the circle $x^2 + y^2 + 8x - 10y + 5 = 0$ ? Give an ordered pair.

Enter a point like (2, -3)