The Equation of a Circle

A circle is a distance

A circle is the set of points at a fixed distance $r$ from a fixed point, its center. That definition already contains the equation — it only has to be written in coordinates.

Put the center at $(h, k)$ and take any point $(x, y)$ on the circle. The horizontal distance between them is $x - h$, the vertical distance is $y - k$, and the straight-line distance is $r$.

Those three lengths form a right triangle, so the Pythagorean theorem applies:

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

Every point on the circle satisfies it, and every point that satisfies it is on the circle — because the equation says exactly that the point is $r$ away from $(h, k)$. That two-way match is what makes it the equation of the circle.

This circle has center $(1, -2)$ and radius $3$:

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

Reading the center and radius

Equation	Center	Radius
$(x - 3)^2 + (y - 5)^2 = 16$	$(3, 5)$	$4$
$(x + 2)^2 + (y - 1)^2 = 49$	$(-2, 1)$	$7$
$x^2 + y^2 = 25$	$(0, 0)$	$5$
$(x - 4)^2 + y^2 = 10$	$(4, 0)$	$\sqrt{10}$

Two traps sit in this table.

The form subtracts $h$ and $k$, so a plus sign means a negative coordinate: $(x + 2)^2$ is $(x - (-2))^2$, giving $h = -2$.

The right side is $r^2$, not $r$. Take the square root to get the radius — an equation ending in $49$ has radius $7$.

Why the signs flip

The equation measures the distance from $(x, y)$ to the center. Distance from the center means subtracting the center’s coordinates, so they appear with a minus.

A center at $x = -2$ is subtracted as $x - (-2)$, which simplifies to $x + 2$. The plus sign is a double negative that has been tidied up, which is the same reason vertex form shows its vertex with flipped signs.

Checking a point

Substitute the point and compare with $r^2$.

Is $(4, 2)$ on the circle $(x - 1)^2 + (y + 2)^2 = 25$?

$$(4 - 1)^2 + (2 + 2)^2 = 9 + 16 = 25 \quad\checkmark$$

Yes. The comparison tells you more when it fails:

Left side compared with $r^2$	The point is
equal	on the circle
less	inside
greater	outside

Expanded form and completing the square

Multiply out $(x - 3)^2 + (y + 1)^2 = 4$ and collect everything on one side:

$$x^2 + y^2 - 6x + 2y + 6 = 0$$

It is the same circle, and neither the center nor the radius is visible anymore. To recover them, complete the square once in $x$ and once in $y$.

Find the center and radius of $x^2 + y^2 + 8x - 4y - 5 = 0$.

1. Group and move the constant.

$$(x^2 + 8x) + (y^2 - 4y) = 5$$

2. Complete each square. Half of $8$ is $4$, and $4^2 = 16$. Half of $-4$ is $-2$, and $(-2)^2 = 4$. Add both to both sides.

$$(x^2 + 8x + 16) + (y^2 - 4y + 4) = 5 + 16 + 4$$

3. Factor.

$$(x + 4)^2 + (y - 2)^2 = 25$$

Center $(-4, 2)$, radius $5$.

Worked examples

Common mistakes

Practice problems