Why is the right side r^2 instead of r?

Because the equation comes from squaring both sides of the distance formula. Starting from sqrt((x-h)^2+(y-k)^2)=r, squaring removes the radical and turns the right side into r^2. So whenever a problem gives you a radius, square it before writing the equation, and whenever an equation gives you the right-hand number, take its square root to get the radius.

How do I tell whether an equation is a circle at all?

In general form, the x^2 and y^2 terms must both be present, have the same sign, and have equal coefficients, and there can be no xy term. If the coefficients differ, you get an ellipse or hyperbola instead. Even when it looks like a circle, complete the square and check the right side: positive gives a real circle, zero gives a single point, and negative gives no graph.

What if the equation has a plus sign, like (x+3)^2?

Rewrite it as subtraction of a negative: (x+3)^2=(x-(-3))^2, so h=-3. The center coordinate is always the opposite of the number you see inside the parentheses. Doing this rewrite on paper for the first several problems will break the habit of reporting the sign backwards.

Do I need the midpoint formula for these problems?

Only when a circle is described by the endpoints of a diameter. In that case the center is the midpoint of those endpoints, and the radius is the distance from that midpoint to either endpoint. If the problem gives you the center directly along with a point on the circle, you only need the distance formula.

GEOM-9.4

Equations of Circles in the Coordinate Plane

Learn to derive the circle equation from the distance formula, read center and radius from standard form, and complete the square to convert general form.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Equations of Circles in the Coordinate Plane, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start Equations of Circles in the Coordinate Plane free

What this lesson covers

Every point on a circle sits exactly the same distance from the center. That single sentence, combined with the distance formula you learned in coordinate geometry, is enough to build the equation of any circle in the plane — no new tools required. In this lesson you will see where the standard form

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

comes from, pull the center and radius straight out of it, and build equations from a center and a point, or from the two endpoints of a diameter.

Then you will run the process backwards. Circles often show up written as

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

, which hides the center completely. Completing the square rewrites that general form into standard form so the geometry becomes visible again. The sign errors that trip people up here are predictable, so we will name them before they cost you anything.

Building the Equation from the Distance Formula

A circle is the set of all points a fixed distance

r

from a fixed center point. Call the center

(h,k)

and let

(x,y)

be any point on the circle. The distance formula says

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

Squaring both sides clears the radical and gives the standard form of a circle:

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

That is the whole derivation, and your teacher may ask you to reproduce it. Notice that squaring is safe here because both sides are nonnegative — a distance and a radius are never negative — so no solutions are gained or lost.

A useful way to picture it: the expression

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

is a machine that measures squared distance from the center. Setting it equal to

r^2

selects exactly the points on the circle. Setting it less than

r^2

describes the interior, and greater than

r^2

describes the exterior. That gives you a fast test for whether a point lies on, inside, or outside a circle: substitute the coordinates and compare the result to

r^2

.

When the center is the origin,

h=k=0

and the equation collapses to

x^2+y^2=r^2

. This is the same relationship as the Pythagorean Theorem, which is not a coincidence: the horizontal and vertical legs from the center to a point on the circle form a right triangle whose hypotenuse is the radius.

Reading Center and Radius from Standard Form

Standard form is written with subtraction inside each set of parentheses, so the center coordinates are the opposites of the numbers you see. The right side is

r^2

, not

r

, so you must take a square root to get the radius.

Equation	Center	Radius
$(x-3)^2+(y-8)^2=49$	$(3,8)$	$7$
$(x+5)^2+(y-1)^2=20$	$(-5,1)$	$2\sqrt{5}$
$x^2+(y+6)^2=1$	$(0,-6)$	$1$
$(x-2)^2+y^2=12$	$(2,0)$	$2\sqrt{3}$

To handle a plus sign, rewrite it as subtraction of a negative:

(x+5)^2=(x-(-5))^2

, so

h=-5

. Students who skip this step reliably report the center as

(5,1)

instead of

(-5,1)

.

The second common slip is reporting the radius as the number on the right. In

(x+5)^2+(y-1)^2=20

, the radius is

\sqrt{20}=2\sqrt{5}\approx 4.47

, not 20. Leave irrational radii in simplified radical form unless the problem asks for a decimal.

A missing variable term means that coordinate of the center is zero, not that the term is missing from the circle. In

x^2+(y+6)^2=1

, the

x

-term is really

(x-0)^2

.

To graph from standard form, plot the center, then step

r

units right, left, up, and down to get four points on the circle, and sketch a smooth curve through them.

Writing Equations from Geometric Descriptions

Most problems hand you the geometry and ask for the algebra. Two setups appear constantly.

Center and a point on the circle. The radius is the distance from the center to that point, so compute

r^2

directly with the distance formula and never bother taking a square root —

r^2

is what the equation needs. If the center is

(1,-2)

and the circle passes through

(5,1)

, then

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

, and the equation is

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

.

Endpoints of a diameter. The center is the midpoint of the diameter, and the radius is half the diameter's length. For endpoints

A

and

B

, use

\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)

for the center, then find

r^2

as the squared distance from that center to either endpoint. Finding the full diameter length and forgetting to halve it is the single most common mistake here; the resulting circle is four times too large in area.

Tangent to an axis. If a circle with center

(h,k)

is tangent to the

x

-axis, the shortest distance from the center to that line is the radius, so

r=|k|

. Tangent to the

y

-axis gives

r=|h|

. Sketching the situation makes this obvious and prevents guessing.

Always finish by checking one known point. Substitute the given point back into your equation; if the two sides are not equal, an arithmetic slip happened somewhere in the radius computation.

Completing the Square: General Form to Standard Form

Expanding standard form produces the general form

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

, in which the center and radius are invisible. Completing the square reverses the expansion.

The procedure, for

x^2+y^2-6x+10y+18=0

Step	Result
Group $x$ terms and $y$ terms, move the constant	$(x^2-6x)+(y^2+10y)=-18$
Half of $-6$ is $-3$ ; square it to get $9$	add $9$ to both sides
Half of $10$ is $5$ ; square it to get $25$	add $25$ to both sides
Factor each perfect-square trinomial	$(x-3)^2+(y+5)^2=16$

Center

(3,-5)

, radius

4

.

The most common error is adding the completing-the-square constants to only the left side. Whatever you add inside the parentheses must also be added to the right side, or the equation is no longer equivalent.

If the squared terms have a common coefficient other than 1 — for example

3x^2+3y^2-12x+6y=0

— divide every term by that coefficient first, then complete the square. (If the coefficients on

x^2

and

y^2

are different, the graph is not a circle at all.)

Finally, interpret the right side. If it comes out positive, you have a genuine circle. If it equals zero, the graph is a single point, sometimes called a degenerate circle. If it is negative, no real points satisfy the equation and the graph is empty — that is a legitimate answer, not a sign that you made a mistake, though it is worth rechecking your arithmetic.

Where Students Actually Go Wrong

Four errors account for most lost work on this topic, and all four are avoidable with a habit.

First, sign flips on the center. The form is

(x-h)^2

, so the center coordinate is the opposite of the constant you read. Write

(x+7)^2

(x-(-7))^2

on scratch paper the first few times.

Second, confusing $r$ and $r^2$ . If a problem gives a radius of

\sqrt{10}

, the right side is

10

. If a problem gives a radius of

6

, the right side is

36

. Ask yourself which one you were handed before writing anything.

Third, halving mistakes with diameters. A diameter of length

10

gives

r=5

and

r^2=25

, not

r^2=100

.

Fourth, incomplete balancing when completing the square. Track the added constants explicitly on the right side as you go.

A reliable self-check: pick a point that should be on the circle and substitute. For

(x-3)^2+(y+5)^2=16

with center

(3,-5)

and radius

4

, the point

(7,-5)

should work:

(7-3)^2+(-5+5)^2=16+0=16

. It checks.

This coordinate description of circles connects back to the chord, arc, and tangent relationships from earlier in the unit. Once you can write a circle's equation, you can find where a line meets it by substitution, verify that a radius drawn to a tangent point is perpendicular to the tangent line, or confirm that a given point lies inside or outside a region.

Key terms

Circle.: The set of all points in a plane at a fixed distance (the radius) from a fixed point (the center).
Standard form of a circle.: $(x-h)^2+(y-k)^2=r^2$ , where $(h,k)$ is the center and $r$ is the radius.
General form of a circle.: $x^2+y^2+Dx+Ey+F=0$ ; an expanded version of standard form in which the center and radius are not directly visible.
Completing the square.: Adding the square of half the linear coefficient to both sides of an equation to create a perfect-square trinomial such as $x^2-6x+9=(x-3)^2$ .
Radius.: The distance from the center of a circle to any point on the circle; it equals half the diameter.
Diameter.: A chord passing through the center; its midpoint is the center and its length is $2r$ .
Degenerate circle.: The graph of an equation in standard form whose right side is zero, which is a single point rather than a curve.
Distance formula.: $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ , the algebraic statement of the Pythagorean Theorem used to derive the circle equation.

Worked example

Convert

2x^2+2y^2+8x-20y-10=0

to standard form. State the center and radius, then determine whether the point

(-2,1)

lies inside, on, or outside the circle.

Step 1: The coefficients on

x^2

and

y^2

match but are not 1, so divide every term by 2:

x^2+y^2+4x-10y-5=0

.

Step 2: Group like variables and move the constant to the right:

(x^2+4x)+(y^2-10y)=5

.

Step 3: Complete the square for

x

. Half of

4

2

, and

2^2=4

. Add

4

to both sides.

Step 4: Complete the square for

y

. Half of

-10

-5

, and

(-5)^2=25

. Add

25

to both sides. The right side becomes

5+4+25=34

.

Step 5: Factor:

(x+2)^2+(y-5)^2=34

.

Step 6: Read the answers. Center is

(-2,5)

because

(x+2)^2=(x-(-2))^2

. Radius is

\sqrt{34}\approx 5.83

, since the right side is

r^2

, not

r

.

Step 7: Test the point

(-2,1)

. Substitute into the left side:

(-2+2)^2+(1-5)^2=0+16=16

. Compare to

r^2=34

. Since

16<34

, the squared distance from the center is less than the squared radius, so

(-2,1)

lies inside the circle.

Quick sanity check: the point

(-2,1)

is directly below the center

(-2,5)

by 4 units, and 4 is less than

\sqrt{34}\approx 5.83

, so it should indeed be inside.

Practice questions

Which equation represents the circle with center

(-4,7)

and radius

3

$(x+4)^2+(y-7)^2=9$
$(x-4)^2+(y+7)^2=9$
$(x+4)^2+(y-7)^2=3$
$(x-4)^2+(y+7)^2=3$

Answer: $(x+4)^2+(y-7)^2=9$

Standard form subtracts the center coordinates:

(x-(-4))^2+(y-7)^2

simplifies to

(x+4)^2+(y-7)^2

. The right side must be

r^2=3^2=9

. Choices with

(x-4)

and

(y+7)

flip both signs of the center, and choices ending in 3 use the radius where

r^2

belongs.

A circle has a diameter with endpoints

A(-2,3)

and

B(6,9)

. Write the equation of the circle in standard form and state its center and radius.

Answer: $(x-2)^2+(y-6)^2=25$ , with center $(2,6)$ and radius $5$ .

The center is the midpoint of the diameter:

\left(\frac{-2+6}{2},\frac{3+9}{2}\right)=(2,6)

. For the radius, find the distance from the center to either endpoint: from

(2,6)

(6,9)

the horizontal change is 4 and the vertical change is 3, so

r^2=16+9=25

and

r=5

. (Equivalently, the full diameter is

\sqrt{64+36}=10

, so the radius is 5 — the halving step is where most errors occur.) Substituting gives

(x-2)^2+(y-6)^2=25

Convert

x^2+y^2+2x-8y+17=0

to standard form and describe its graph.

Answer: $(x+1)^2+(y-4)^2=0$ ; the graph is the single point $(-1,4)$ .

Group and move the constant:

(x^2+2x)+(y^2-8y)=-17

. Half of 2 is 1, so add

1

; half of

-8

-4

, so add

16

. The right side becomes

-17+1+16=0

, giving

(x+1)^2+(y-4)^2=0

. Since squares cannot be negative, the only solution is

x=-1

and

y=4

. This is a degenerate circle with radius zero — a single point, not an error in your work.

FAQ

Why is the right side $r^2$ instead of $r$ ?: Because the equation comes from squaring both sides of the distance formula. Starting from $\sqrt{(x-h)^2+(y-k)^2}=r$ , squaring removes the radical and turns the right side into $r^2$ . So whenever a problem gives you a radius, square it before writing the equation, and whenever an equation gives you the right-hand number, take its square root to get the radius.
How do I tell whether an equation is a circle at all?: In general form, the $x^2$ and $y^2$ terms must both be present, have the same sign, and have equal coefficients, and there can be no $xy$ term. If the coefficients differ, you get an ellipse or hyperbola instead. Even when it looks like a circle, complete the square and check the right side: positive gives a real circle, zero gives a single point, and negative gives no graph.
What if the equation has a plus sign, like $(x+3)^2$ ?: Rewrite it as subtraction of a negative: $(x+3)^2=(x-(-3))^2$ , so $h=-3$ . The center coordinate is always the opposite of the number you see inside the parentheses. Doing this rewrite on paper for the first several problems will break the habit of reporting the sign backwards.
Do I need the midpoint formula for these problems?: Only when a circle is described by the endpoints of a diameter. In that case the center is the midpoint of those endpoints, and the radius is the distance from that midpoint to either endpoint. If the problem gives you the center directly along with a point on the circle, you only need the distance formula.

Learn this with a teacher, not a page

The Crimsora tutor teaches Equations of Circles in the Coordinate Plane live — explaining on a whiteboard, asking you questions, and adapting to where you get stuck.

Try a live session free