The Equation of a Circle

Learning goals

Measure distance with $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
Define a circle as points a fixed distance from a center
Write standard form $(x - h)^2 + (y - k)^2 = r^2$
Read the center and radius carefully, minding the signs
Complete the square on each variable to leave general form
Recognize a point or no graph from the right side

Measuring distance in the plane

On a number line the distance between two numbers is the size of their difference: the points $2$ and $9$ are $|9 - 2| = 7$ apart. In the plane a point needs two coordinates, so the two points no longer sit on one straight track, and a plain subtraction cannot reach across the gap. What saves you is that any two points are joined by a right triangle hiding in the grid.

Take $(x_1, y_1)$ and $(x_2, y_2)$ . Travel from the first to the second in two steps that meet at a right angle: slide straight across to the corner $(x_2, y_1)$ , then straight up to $(x_2, y_2)$ . The across step spans the gap between the $x$ -coordinates, and the up step spans the gap between the $y$ -coordinates. Those two steps are the legs of a right triangle, and the direct path you actually want, the straight distance, is its hypotenuse. That is exactly the setup the Pythagorean theorem was built for.

The straight-line distance between two points is the hypotenuse of a right triangle. Going from A at negative 2, negative 1 across to a corner and then up to B at 2, 2 gives a horizontal leg of length 4 and a vertical leg of length 3, so the distance is the square root of 4 squared plus 3 squared, which equals 5.

Why the distance is $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ #

Place the two points $(x_1, y_1)$ and $(x_2, y_2)$ on the grid, and drop to the corner $(x_2, y_1)$ . That corner shares its height with the first point and its side-to-side position with the second. The horizontal leg runs from $(x_1, y_1)$ to that corner, so its length is the gap between the $x$ -coordinates, $|x_2 - x_1|$ . The vertical leg runs from the corner up to $(x_2, y_2)$ , so its length is the gap between the $y$ -coordinates, $|y_2 - y_1|$ . These legs meet at a right angle, and the distance $d$ you want is the hypotenuse.

The Pythagorean theorem says the square of the hypotenuse equals the sum of the squares of the two legs:

d^2 = |x_2 - x_1|^2 + |y_2 - y_1|^2.

Squaring a number throws away its sign, so $|x_2 - x_1|^2$ is simply $(x_2 - x_1)^2$ , and the same holds for the vertical leg. The absolute-value bars are no longer doing anything:

d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2.

A distance is never negative, so take the positive square root of both sides:

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.

Because each difference is squared, it makes no difference which point you list first. Swapping them flips the sign inside each set of parentheses, and squaring erases that flip, so the distance comes out the same either way.

This result is the distance formula, and it is nothing more than the Pythagorean theorem applied to the legs you can read off the coordinates.

Worked example 1 Find the distance between $(-1, 2)$ and $(3, 5)$

Label the points $(x_1, y_1) = (-1, 2)$ and $(x_2, y_2) = (3, 5)$ , and substitute into the distance formula. The horizontal gap is $3 - (-1) = 4$ and the vertical gap is $5 - 2 = 3$ :

d = \sqrt{(3 - (-1))^2 + (5 - 2)^2} = \sqrt{4^2 + 3^2}.

Now square and add before taking the root, since $4^2 + 3^2 = 16 + 9 = 25$ :

d = \sqrt{25} = 5.

The two points are $5$ units apart. It is the same $3$ - $4$ - $5$ right triangle you would get by counting $4$ across and $3$ up between them.

Check your understanding

How far apart are the points $(0, 0)$ and $(6, 8)$ ?

10

14

\sqrt{14}

100

What a circle is

Now that distance is under control, define the shape precisely. A circle is the set of all points in the plane that lie a fixed distance $r$ from a fixed point. The fixed point is the center, usually written $(h, k)$ , and the fixed distance $r$ (a positive number) is the radius. Every point on the circle is exactly $r$ from the center, and every point that is exactly $r$ from the center is on the circle. Nothing else qualifies.

That definition is stated entirely in terms of distance, which is why the distance formula is the key that unlocks the equation. The equation of a circle is just the sentence “this point is $r$ away from the center” written in algebra.

The standard equation of a circle

Let $(x, y)$ be a point anywhere on a circle with center $(h, k)$ and radius $r$ . Saying $(x, y)$ is on the circle means its distance from the center equals $r$ . Feed the center and the point into the distance formula and set the result equal to $r$ , then clear the square root by squaring both sides.

A point x, y on a circle of center h, k and radius r sits at the end of a right triangle. The horizontal leg is x minus h, the vertical leg is y minus k, and the radius r is the hypotenuse, so the Pythagorean theorem gives x minus h squared plus y minus k squared equals r squared.

Why a circle's equation is $(x - h)^2 + (y - k)^2 = r^2$ #

A point $(x, y)$ lies on the circle exactly when its distance from the center $(h, k)$ equals the radius $r$ . Measure that distance with the formula just derived, treating $(h, k)$ as one point and $(x, y)$ as the other:

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

This already describes the circle, but the square root is clumsy to work with. Remove it by squaring both sides. The left side is a distance and the right side is a positive radius, so both sides are nonnegative. Since both are nonnegative, squaring keeps the equation equivalent, neither creating nor destroying any solutions:

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

Every step above reverses, so the statement runs in both directions. A point $(x, y)$ satisfies this equation if and only if it sits exactly $r$ from $(h, k)$ , which is precisely what it means to be on the circle. That is why this one equation captures the entire curve and no other points.

This is the standard form, also called the center-radius form, of a circle:

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

Its power is that the center and radius sit right on the surface. The numbers subtracted from $x$ and $y$ are the center coordinates, and the number on the right is the radius squared.

Circles centered at the origin

The simplest circles are centered at the origin, where $h = 0$ and $k = 0$ . Substituting those into the standard form makes the subtractions vanish:

(x - 0)^2 + (y - 0)^2 = r^2 \quad\Longrightarrow\quad x^2 + y^2 = r^2.

So a circle centered at the origin with radius $r$ has the tidy equation $x^2 + y^2 = r^2$ . Here the right side is $r^2$ , so the radius is the square root of it: $x^2 + y^2 = 9$ has radius $\sqrt{9} = 3$ , not $9$ . This circle meets each axis exactly $r$ from the origin, at $(r, 0)$ , $(-r, 0)$ , $(0, r)$ , and $(0, -r)$ .

A circle centered at the origin with radius 3 has equation x squared plus y squared equals 9. It crosses the axes exactly r units from the center, at the four points 3 and 0, negative 3 and 0, 0 and 3, and 0 and negative 3.

Reading the center and radius

The standard form works in both directions. Going forward, you build the equation from a center and a radius; going backward, you read a center and a radius out of a finished equation. Two small traps trip people up in the backward direction, and both come from taking the form literally.

The sign trap comes from the subtraction built into $x - h$ and $y - k$ . A circle written with $(x + 3)$ does not have center coordinate $3$ . Rewrite $x + 3$ as $x - (-3)$ to see that $h = -3$ . The form always subtracts the center coordinate, so a plus sign inside the parentheses means that coordinate is negative.

The radius trap comes from the right side being $r^2$ , not $r$ . To get the radius you must take the square root of the number on the right. In $(x - 1)^2 + (y - 2)^2 = 16$ the radius is $\sqrt{16} = 4$ , and when the right side is not a perfect square, such as $20$ , the radius stays as a simplified root like $\sqrt{20} = 2\sqrt{5}$ .

Worked example 2 Write the equation of the circle with center $(-3, 4)$ and radius $6$

Here $h = -3$ , $k = 4$ , and $r = 6$ . Substitute into $(x - h)^2 + (y - k)^2 = r^2$ , watching the signs. Subtracting the negative $h$ turns into adding:

(x - (-3))^2 + (y - 4)^2 = 6^2.

Simplify $x - (-3)$ to $x + 3$ , and $6^2$ to $36$ :

(x + 3)^2 + (y - 4)^2 = 36.

The $+3$ inside the first set of parentheses is the fingerprint of a center whose $x$ -coordinate is $-3$ . The form subtracts the center coordinate, so a plus sign inside always signals a negative coordinate.

Worked example 3 Find the center and radius of $(x - 5)^2 + (y + 2)^2 = 20$

Match the equation against the standard form $(x - h)^2 + (y - k)^2 = r^2$ one piece at a time. The first parenthesis is $x - 5$ , so $h = 5$ . The second is $y + 2$ , which is $y - (-2)$ , so $k = -2$ . The center is therefore $(5, -2)$ .

The right side is $r^2 = 20$ , so the radius is its square root, not $20$ itself:

r = \sqrt{20} = \sqrt{4 \cdot 5} = 2\sqrt{5}.

The center is $(5, -2)$ and the radius is $2\sqrt{5}$ . Both traps appear in this one problem: the $+2$ makes $k$ negative, and the radius is $\sqrt{20}$ , not the $20$ printed on the right.

Check your understanding

What are the center and radius of the circle $(x + 4)^2 + (y - 3)^2 = 25$ ?

A center

(4, -3)

, radius

5

B center

(-4, 3)

, radius

25

C center

(-4, 3)

, radius

5

D center

(4, -3)

, radius

25

The radius is just the distance from the center to any point on the circle. So the distance formula lets you build a circle’s equation even when you are handed a point on the circle instead of the radius.

Worked example 4 Write the equation of the circle centered at $(2, 1)$ that passes through $(5, 5)$

The radius is the distance from the center to any point on the circle, so it is the distance from $(2, 1)$ to $(5, 5)$ . In fact you can skip the square root entirely, because the equation only ever needs $r^2$ , and the distance formula gives $r^2$ before you take any root:

r^2 = (5 - 2)^2 + (5 - 1)^2 = 3^2 + 4^2 = 9 + 16 = 25.

Now place the center $(2, 1)$ and this $r^2 = 25$ straight into the standard form:

(x - 2)^2 + (y - 1)^2 = 25.

The circle has radius $\sqrt{25} = 5$ , but you never had to compute that root. Reading $r^2$ directly off the distance calculation is quicker and avoids a needless square-root-then-square round trip.

The general form

Multiply out a circle’s standard form and it stops looking like a circle. Expanding $(x - h)^2 + (y - k)^2 = r^2$ gives $x^2 - 2hx + h^2 + y^2 - 2ky + k^2 = r^2$ , and moving everything to one side collects into

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

where $D = -2h$ , $E = -2k$ , and $F = h^2 + k^2 - r^2$ . This is the general form. It is a genuine circle in disguise, but the center and radius are buried, and you cannot read them off directly.

To dig them back out, reverse the expansion by completing the square on the $x$ -terms and on the $y$ -terms separately. That is the same move you learned in the completing-the-square lesson. Group the $x$ -terms, group the $y$ -terms, and turn each group into a perfect square. Once the equation is back in standard form, the center and radius reappear.

Worked example 5 Convert $x^2 + y^2 - 6x + 8y - 11 = 0$ to standard form

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

(x^2 - 6x) + (y^2 + 8y) = 11.

Complete the square on each group separately. For $x^2 - 6x$ , half of $-6$ is $-3$ and $(-3)^2 = 9$ . For $y^2 + 8y$ , half of $8$ is $4$ and $4^2 = 16$ . Add each of those constants to both sides so the equation stays balanced:

(x^2 - 6x + 9) + (y^2 + 8y + 16) = 11 + 9 + 16.

Each group is now a perfect square, and the right side totals $36$ :

(x - 3)^2 + (y + 4)^2 = 36.

Read it off: the center is $(3, -4)$ , from $x - 3$ and $y + 4$ , and the radius is $\sqrt{36} = 6$ . What looked like a scrambled general form was a clean circle all along.

Check your understanding

Convert $x^2 + y^2 + 10x - 4y + 20 = 0$ to standard form. What are its center and radius?

A center

(5, -2)

, radius

3

B center

(-5, 2)

, radius

3

C center

(-5, 2)

, radius

9

D center

(-5, 2)

, radius

\sqrt{29}

When the circle collapses

Completing the square on a general form always lands you at $(x - h)^2 + (y - k)^2 = R$ for some number $R$ on the right. Usually $R$ is positive and you have an honest circle of radius $\sqrt{R}$ . But the right side is not guaranteed to be positive, and two special cases are worth recognizing.

If $R = 0$ , the equation reads $(x - h)^2 + (y - k)^2 = 0$ . A sum of two squares can be zero only when each square is zero, which forces $x = h$ and $y = k$ . The “circle” has shrunk to the single point $(h, k)$ . For instance $x^2 + y^2 - 2x + 4y + 5 = 0$ becomes $(x - 1)^2 + (y + 2)^2 = 0$ , satisfied only by the point $(1, -2)$ .

If $R < 0$ , the equation asks for a sum of two squares to be negative. Squares are never negative, so no real point $(x, y)$ can satisfy it, and the graph is empty: there is no curve at all. For instance $x^2 + y^2 - 2x + 4y + 8 = 0$ becomes $(x - 1)^2 + (y + 2)^2 = -3$ , which no real point solves.

A true circle needs $R > 0$ . Always check the sign of the right side after completing the square before you announce a radius.

Graphing a circle

Once you know the center and radius, a circle is quick to draw by hand. Plot the center, then step out the radius $r$ in each of the four cardinal directions. That marks four points: right to $(h + r, k)$ , left to $(h - r, k)$ , up to $(h, k + r)$ , and down to $(h, k - r)$ . Those four anchors pin the circle in place, and a smooth round curve through them completes it.

Worked example 6 Graph $(x - 1)^2 + (y + 1)^2 = 9$

First read the center and radius. The parentheses are $x - 1$ and $y + 1 = y - (-1)$ , so the center is $(1, -1)$ . The right side is $9 = r^2$ , so the radius is $\sqrt{9} = 3$ .

Plot the center at $(1, -1)$ , then step out $3$ units in each direction to reach four points of the circle:

\begin{aligned} (1 + 3,\, -1) &= (4, -1), &\qquad (1 - 3,\, -1) &= (-2, -1), \\ (1,\, -1 + 3) &= (1, 2), &\qquad (1,\, -1 - 3) &= (1, -4). \end{aligned}

Sketch a round curve through those four anchor points, and the circle is done.

To graph x minus 1 squared plus y plus 1 squared equals 9, plot the center at 1, negative 1 and step out the radius 3 in four directions to reach 4, negative 1 and negative 2, negative 1 and 1, 2 and 1, negative 4. A round curve through those four points is the circle.

Sometimes you are handed the two endpoints of a diameter instead of the center. The center is the midpoint of that diameter, exactly halfway between the endpoints, so average their $x$ -coordinates and average their $y$ -coordinates to locate it. The radius is then the distance from the center to either endpoint.

Common mistakes

Flipping the sign of the center. The form is $(x - h)^2 + (y - k)^2 = r^2$ , so it subtracts the center coordinate. A circle with $(x + 3)$ has $h = -3$ , not $+3$ . Rewrite $x + 3$ as $x - (-3)$ to keep the sign straight.
Using the right side as the radius. The right side is $r^2$ . The radius is its square root, so $(x - 1)^2 + (y - 2)^2 = 16$ has radius $4$ , not $16$ , and a right side of $20$ gives radius $\sqrt{20} = 2\sqrt{5}$ , not $20$ .
Forgetting the square root in the distance formula. The formula ends in a square root: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ . Stopping at $(x_2 - x_1)^2 + (y_2 - y_1)^2$ gives $d^2$ , not $d$ .
Adding coordinates instead of squaring differences. Distance is not $|x_2 - x_1| + |y_2 - y_1|$ . You square each difference, add, then take the root, because the two gaps are the legs of a right triangle.
Completing the square on only one variable. The general form needs the square completed on the $x$ -terms and on the $y$ -terms, each with its own constant. Both constants must then be added to the right side.
Announcing a circle without checking the right side. After completing the square, if the right side is $0$ the graph is a single point, and if it is negative there is no graph at all. Only a positive right side gives a real circle.
Ignoring a leading coefficient on $x^2$ and $y^2$ . For an equation to be a circle, the $x^2$ and $y^2$ terms must have equal coefficients. If those coefficients are equal but not $1$ (say both $2$ ), first divide the whole equation by that common value so both become $1$ , then complete the square. If the coefficients are unequal, the equation is not a circle at all.

Key takeaways

The distance between $(x_1, y_1)$ and $(x_2, y_2)$ is $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ , which is the Pythagorean theorem on the horizontal and vertical gaps.
A circle is the set of points a fixed distance $r$ (the radius) from a fixed center $(h, k)$ .
Writing “distance from $(x, y)$ to $(h, k)$ equals $r$ ” and squaring gives the standard form $(x - h)^2 + (y - k)^2 = r^2$ ; centered at the origin it becomes $x^2 + y^2 = r^2$ .
Read it off carefully: the center is $(h, k)$ where the form subtracts $h$ and $k$ (so $x + 3$ means $h = -3$ ). The radius is $\sqrt{\text{right side}}$ , not the right side itself.
The general form $x^2 + y^2 + Dx + Ey + F = 0$ becomes standard form by completing the square on the $x$ -terms and the $y$ -terms separately, then reading the center and radius.
After completing the square, a right side of $0$ gives a single point and a negative right side gives no graph. A real circle needs a positive right side.
To graph a circle, plot the center and step out the radius right, left, up, and down to four anchor points, then draw a smooth curve through them.

The Equation of a Circle

Learning goals

Measuring distance in the plane

What a circle is

The standard equation of a circle

Circles centered at the origin

Reading the center and radius

The general form

When the circle collapses

Graphing a circle

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)