Circle (standard equation)

A circle looks simple — a round curve traced at a fixed distance from a centre — but that one geometric idea, "constant distance," is enough to generate a full algebraic equation you can manipulate, expand, and reverse-engineer. This is where coordinate geometry meets the oldest curve in geometry: instead of drawing a circle with a compass, you write down exactly which pairs $(x,y)$ satisfy $\sqrt{(x-h)^2+(y-k)^2}=r$ , square it, and get a clean equation you can compare, expand, or complete the square on. This matters because every circle problem in the Conic Sections chapter — finding a circle through given points, checking whether an equation is a circle at all, extracting centre and radius from a messy expanded form — reduces to fluency with two equivalent equations: the standard form $(x-h)^2+(y-k)^2=r^2$ and the general form $x^2+y^2+2gx+2fy+c=0$ . Master converting between them and this whole topic becomes mechanical.

The core insight

A circle is the set of points at constant distance $r$ from a centre $(h,k)$ , so squaring the distance formula gives $(x-h)^2+(y-k)^2=r^2$ , which expands into the general form $x^2+y^2+2gx+2fy+c=0$ with centre $(-g,-f)$ and radius $\sqrt{g^2+f^2-c}$ .

The core idea

A circle is defined purely by a locus condition: a point $P(x,y)$ stays on the circle exactly when its distance from a fixed centre $C(h,k)$ equals a fixed radius $r>0$ . Nothing else about $P$ matters — not its exact position, only that this one distance stays constant. Applying the distance formula to that condition and squaring both sides removes the square root and hands you a polynomial equation in $x$ and $y$ :

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

where: $(h,k)$ is the centre, $r$ is the radius, and $(x,y)$ is any point on the circle. Setting $h=k=0$ gives the centre-at-origin case $x^2+y^2=r^2$ , the simplest circle of all. Expanding the standard form and collecting terms produces the general equation of a circle:

x^2+y^2+2gx+2fy+c=0,

where: $g=-h$ , $f=-k$ , and $c=h^2+k^2-r^2$ . Reading this backwards, any equation of this shape has centre $(-g,-f)$ and radius $\sqrt{g^2+f^2-c}$ , provided $g^2+f^2-c>0$ . If that quantity is exactly $0$ the equation describes a single point, and if it is negative there is no real curve at all — the equation looks like a circle algebraically but has no locus in the real plane. Two structural facts are worth noticing before touching any problem. First, a second-degree equation in $x,y$ is a circle only if the coefficients of $x^2$ and $y^2$ are equal and there is no $xy$ term — an equation like $2x^2+3y^2+\dots=0$ or one with an $xy$ term is never a circle, no matter how it is manipulated. Second, when a circle is given through the endpoints of a diameter rather than its centre, there is a shortcut: if $A(x_1,y_1)$ and $B(x_2,y_2)$ are diameter endpoints, any other point $P$ on the circle sees $AP\perp BP$ (angle in a semicircle), so the slopes multiply to $-1$ , giving

(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0.

This diameter shortcut is worth pausing on because it swaps out the usual centre-and-radius route for a purely point-based one. Instead of first finding the midpoint of $AB$ (which would give the centre) and then the half-distance $|AB|/2$ (which would give the radius), you can plug $A$ and $B$ straight into the product form above and expand. Both routes always agree, since the midpoint of a diameter is by definition the centre and half its length is by definition the radius — the shortcut simply skips the intermediate arithmetic when only the endpoints are given.

Reading centre and radius quickly. Once an equation is in general form, three numbers control everything about the circle: $g$ and $f$ fix where the centre sits (with the sign flipped), and $c$ combines with $g$ and $f$ through $g^2+f^2-c$ to fix how big the circle is. This is why comparing coefficients term by term against $x^2+y^2+2gx+2fy+c=0$ — rather than re-deriving the centre and radius from scratch by completing the square every time — is the fastest way to extract information from an equation already given in expanded form. Completing the square remains useful as a cross-check, and it is the only reliable method when the coefficients of $x^2$ and $y^2$ are not already $1$ and need to be divided out first.

Figure

A circle with centre C at coordinates h k and a point P on the circumference joined to the centre by a radius r, illustrating the distance definition of a circle

Derivation AThe standard equation of a circle from its distance definition

The goal is to turn the purely geometric definition of a circle — the set of points at a fixed distance $r$ from a fixed centre $(h,k)$ — into an algebraic equation in $x$ and $y$ that can be expanded, compared, and manipulated like any other equation, using nothing beyond the distance formula and squaring both sides.

Step 1. Write the defining condition. A point $P(x,y)$ lies on the circle exactly when its distance from the centre $C(h,k)$ equals $r$ . By the distance formula between two points, this distance is $\sqrt{(x-h)^2+(y-k)^2}$ , so the condition is $\sqrt{(x-h)^2+(y-k)^2}=r$ .

Step 2. Square both sides to remove the square root. Since $r>0$ and a distance is never negative, squaring is a reversible step here (no extraneous solutions are introduced), giving $(x-h)^2+(y-k)^2=r^2$ — the standard equation of a circle.

Step 3. Expand to reach the general form. Expanding both squared terms, $x^2-2hx+h^2+y^2-2ky+k^2=r^2$ , then collecting: $x^2+y^2-2hx-2ky+(h^2+k^2-r^2)=0$ . Substituting $g=-h$ , $f=-k$ , $c=h^2+k^2-r^2$ gives $x^2+y^2+2gx+2fy+c=0$ .

Step 4. Reverse the process to recover centre and radius from a general-form equation. Given $x^2+y^2+2gx+2fy+c=0$ , completing the square on the $x$ -terms and $y$ -terms separately gives $(x+g)^2+(y+f)^2=g^2+f^2-c$ , which matches the standard form with centre $(-g,-f)$ and radius $\sqrt{g^2+f^2-c}$ .

Worked example

Problem. Find the equation of the circle with centre $(2,-3)$ and radius $5$ , then verify by extracting the centre and radius back from the general form.

Step 1. Substitute $h=2$ , $k=-3$ , $r=5$ into the standard form: $(x-h)^2+(y-k)^2=r^2$ gives $(x-2)^2+(y+3)^2=25$ .

Step 2. Expand to reach general form: $x^2-4x+4+y^2+6y+9=25$ , which simplifies to $x^2+y^2-4x+6y-12=0$ .

Step 3. Verify by comparing with $x^2+y^2+2gx+2fy+c=0$ : here $2g=-4\Rightarrow g=-2$ and $2f=6\Rightarrow f=3$ , so centre $=(-g,-f)=(2,-3)$ , matching the given centre, and radius $=\sqrt{g^2+f^2-c}=\sqrt{4+9+12}=\sqrt{25}=5$ , matching the given radius.

Answer: standard form $(x-2)^2+(y+3)^2=25$ ; general form $x^2+y^2-4x+6y-12=0$ ; both checks confirm the same circle.

Competitive stretch

Problem. Find the centre and radius of $x^2+y^2-6x+8y-11=0$ .

Step 1. Compare with $x^2+y^2+2gx+2fy+c=0$ : $2g=-6\Rightarrow g=-3$ ; $2f=8\Rightarrow f=4$ ; $c=-11$ .

Step 2. Centre $=(-g,-f)=(3,-4)$ , and radius $=\sqrt{g^2+f^2-c}=\sqrt{9+16+11}=\sqrt{36}=6$ .

Step 3. Cross-check by completing the square directly: $(x-3)^2-9+(y+4)^2-16=11\Rightarrow(x-3)^2+(y+4)^2=36$ , confirming centre $(3,-4)$ , radius $6$ .

Answer: centre $(3,-4)$ , radius $6$ .

Common traps

Reading the centre off the general-form coefficients as $(g,f)$ instead of $(-g,-f)$ — the sign flip is easy to miss under time pressure.
Dropping or mishandling the constant term while completing the square, which shifts the value under the square root for the radius and can even make it negative.
Assuming any second-degree equation in $x,y$ is a circle — it needs equal, nonzero coefficients for $x^2$ and $y^2$ and no $xy$ term.
Sign slips when squaring the distance formula with a negative centre coordinate, e.g. writing $(x+2)^2$ instead of $(x-2)^2$ when $h=-2$ .
Trying to memorize the diameter-form equation without linking it to the right-angle-in-a-semicircle property behind it — the formula is easy to misremember under pressure if the geometry is forgotten.
Forgetting to divide through first when the coefficients of $x^2$ and $y^2$ in a general-form equation are not already $1$ — comparing directly with $x^2+y^2+2gx+2fy+c=0$ only works once the leading coefficients match.

Ask the tutor: Not sure whether an equation like $x^2+y^2-4x+9=0$ represents a real circle — ask the AI tutor to complete the square with you and check.

Why does squaring the distance formula not introduce extra solutions here?

Because both sides of $\sqrt{(x-h)^2+(y-k)^2}=r$ are non-negative — a distance is never negative and we are given $r>0$ — squaring a true equation between two non-negative quantities is fully reversible. This is different from squaring an equation like $x=\sqrt{y}$ where you must later check for extraneous roots; here every solution of the squared equation genuinely satisfies the original distance condition, so $(x-h)^2+(y-k)^2=r^2$ is exactly equivalent to the locus definition.

How do I quickly tell if a given second-degree equation is a circle without fully completing the square?

Check two things first. The coefficients of $x^2$ and $y^2$ must be equal (and nonzero) and there must be no $xy$ term — if either fails, it is not a circle at all, and you can stop. If both pass, divide through so the coefficient of $x^2$ and $y^2$ is exactly $1$ , read off $g$ and $f$ as half the coefficients of $x$ and $y$ , and compute $g^2+f^2-c$ . Only if that value is strictly positive is there a real circle; zero gives a single point and negative gives no real locus.

What is the fastest way to remember which sign the centre takes from the general form?

Anchor it to completing the square rather than memorizing a rule. Starting from $x^2+2gx+y^2+2fy+c=0$ , completing the square on the $x$ -terms gives $(x+g)^2-g^2$ , so the bracket is $(x-(-g))$ , meaning the centre's $x$ -coordinate is $-g$ , not $g$ . The same logic gives $-f$ for the $y$ -coordinate. If you ever doubt the sign, redo this one line of algebra rather than trusting recall under exam pressure.

Circle (standard equation)

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads