Equation of a Circle

Geometry is the math of shape, size, and position. You use it to find lengths, angles, areas, and volumes.

Equation of a Circle: the central idea

A circle is the set of points a fixed distance $r$ from a center $(h,k)$. Squaring the distance formula gives $(x-h)^2+(y-k)^2=r^2$.

Words you need

Center

The center of a circle is the fixed point $(h,k)$ that every point on the circle sits the same distance away from, and it appears inside the parentheses with its signs reversed.

Radius

The radius $r$ is that fixed distance from the center out to the circle, and it appears in the equation squared, on the right-hand side.

Diameter

The diameter is the distance straight across the circle through the center, so it is always twice the radius; a circle with $r^2=25$ has radius $5$ and diameter $10$.

Center-radius form

The center-radius form, also called standard form, is $(x-h)^2+(y-k)^2=r^2$, the arrangement that shows the center and radius without any extra work.

General form

The general form is $x^2+y^2+Dx+Ey+F=0$, the fully expanded version of the same circle, which hides the center and radius until you complete the square.

Completing the square

Completing the square is the move that turns $x^2-6x$ into $(x-3)^2-9$ by halving the middle coefficient and squaring it, and it is how a general-form circle is converted back to center-radius form.

What to know before this lesson

Review the distance formula, coordinate signs, squared binomials, and radius versus diameter.

If one of those prerequisites is uncertain, use the Geometry subject guide to locate the earlier concept before memorizing a procedure.

The equation of a circle: a worked example

Every step, with the arithmetic

1. Step 1 - Read the questionFind the circle centered at $(-4,1)$ that passes through the point $(2,9)$. The center is known, but the radius is not.
2. Step 2 - Find the two legs from center to pointSideways gap: $2-(-4)=6$. Up-and-down gap: $9-1=8$.
3. Step 3 - Square the legs to get $r^2$$r^2=6^2+8^2=36+64=100$
4. Step 4 - Take the square root for the radius$r=\sqrt{100}=10$. The center and the point form a $6$-$8$-$10$ right triangle.
5. Step 5 - Drop the center and radius into the form$(x-(-4))^2+(y-1)^2=100$
6. Step 6 - Tidy up the double negative$(x+4)^2+(y-1)^2=100$
7. Step 7 - Test the point you were given$(2+4)^2+(9-1)^2=6^2+8^2=36+64=100$. It sits on the circle.
8. Step 8 - Expand to general form if asked$x^2+8x+16+y^2-2y+1=100$, so $x^2+y^2+8x-2y-83=0$.

In $(x-2)^2+(y+3)^2=25$, the center is $(2,-3)$ because $y+3=y-(-3)$, and the radius is $\sqrt{25}=5$.

Reading the center and radius straight off a circle equation

The step-by-step method for the equation of a circle

1. Read the center by reversing the signs inside the two squared factors.
2. Read the radius as the nonnegative square root of the right-hand constant.
3. To build an equation, substitute the center and radius, then test a point at one horizontal or vertical radius.

How to check your answer

The point $(7,-3)$ lies five units right of the center. Substitution gives $25+0=25$.

Test triangle inequalities, angle sums, units, scale, and whether the result is compatible with the diagram without assuming the drawing is exact.

A mistake that changes the mathematics

The constant on the right is $r^2$, not $r$. A right side of $25$ means radius $5$.

Where you will actually use this

Cell tower coverage

A tower at map point $(3,-2)$ reaches $5$ miles, so its coverage edge is $(x-3)^2+(y+2)^2=25$. A house at $(7,1)$ gives $(7-3)^2+(1+2)^2=16+9=25$, so it sits exactly on the edge of the signal.

Locating an earthquake

Each seismograph knows only how far away the quake was, which draws a circle around that station. Three such circles from three stations cross at one point, and that intersection is the epicenter. This is the same trilateration idea GPS uses with satellites.

Sprinkler and irrigation planning

A sprinkler that throws water $12$ feet covers the disk $(x-h)^2+(y-k)^2=144$. Placing several sprinklers so their circles just overlap, and no dry gaps are left between them, is a circle-equation problem on a field plan.

How the equation of a circle connects to the rest of geometry

• Pythagorean theorem — The circle equation is really the distance formula with both sides squared, and the distance formula is the Pythagorean theorem applied to the sideways and up-and-down gaps between two points.
• The unit circle — The unit circle is the special case $x^2+y^2=1$, where the center is the origin and the radius is $1$, which is why its points can be written as $(\cos\theta,\sin\theta)$.
• Quadratic formula — Completing the square is the same maneuver used to derive the quadratic formula, so practicing it on circles makes the algebra version feel familiar.

Try a transfer problem

Complete the square to rewrite $x^2+y^2-6x+4y-12=0$ in center-radius form.

Show the worked answer

Start with $x^2+y^2-6x+4y-12=0$. Group the $x$ terms and the $y$ terms and move the constant across: $(x^2-6x)+(y^2+4y)=12$. Complete the square on $x$ by halving $-6$ to get $-3$ and squaring it to get $9$, so $x^2-6x=(x-3)^2-9$. Complete the square on $y$ by halving $4$ to get $2$ and squaring it to get $4$, so $y^2+4y=(y+2)^2-4$. Substituting gives $(x-3)^2-9+(y+2)^2-4=12$, then $(x-3)^2+(y+2)^2=12+9+4=25$. The center is $(3,-2)$ and the radius is $\sqrt{25}=5$. Check with the point $(8,-2)$, five units to the right of the center: $(8-3)^2+(-2+2)^2=25+0=25$.

Work without copying the example. When finished, use the relevant focused calculator or formula reference to check the setup and result, then correct the first line where your reasoning changed. When the method feels reliable, move to geometry practice questions.

Questions about the equation of a circle

Why is the sign inside the parentheses flipped?

Because the form is literally $(x-h)^2$. When the center's $x$-coordinate is $h=-4$, substituting gives $(x-(-4))^2$, which tidies into $(x+4)^2$. A plus sign inside means a negative coordinate, and a minus sign inside means a positive one.

What if the right-hand side is zero or negative?

If it is zero, the only point that works is the center itself, which is called a point circle. If it is negative, no real point can satisfy it, because two squares added together can never be less than zero, so the graph is empty.

How do I tell a circle from an ellipse?

A circle needs the $x^2$ and $y^2$ terms to have the same coefficient and the same sign, as in $3x^2+3y^2-12=0$. If the coefficients differ, such as $4x^2+9y^2=36$, the shape is an ellipse instead.

How do I check whether a point is inside or outside a circle?

Plug the point in and compare with $r^2$ instead of setting them equal. For $(x-3)^2+(y+2)^2=25$, the point $(5,-1)$ gives $4+1=5$, which is less than $25$, so it is inside. A result larger than $25$ would be outside, and exactly $25$ would be on the circle.