Equation of a circle

A circle looks like a picture, but algebraically it is just a distance rule: every point on it sits exactly $r$ units from a fixed centre. Write that distance condition with the distance formula and you get the equation of a circle — and depending on what a problem hands you (a centre and radius, two ends of a diameter, or three points on the rim), a different equivalent form is the fast way in. This section, from the Circles topic of ISC Class 11 Coordinate Geometry, covers the standard form, the diameter form, the general form, and the parametric form, and shows how to build a circle's equation from three given points by solving for its coefficients directly. The goal is fluency moving between forms, not memorising four disconnected formulas.

The core insight

A circle is the set of points at fixed distance $r$ from a centre $(h,k)$ , and every one of its algebraic forms — standard, diameter, general, parametric — is just that one distance condition written differently.

The core idea

Standard form. By the very definition of a circle, a point $(x,y)$ lies on the circle of centre $(h,k)$ and radius $r$ exactly when its distance from the centre equals $r$ . Squaring the distance formula gives

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

where: $(h,k)$ is the centre and $r$ is the radius. When the centre sits at the origin this collapses to $x^2+y^2=r^2$ , the simplest case worth recognising on sight.

Diameter form. If $A(x_1,y_1)$ and $B(x_2,y_2)$ are the two ends of a diameter, then for any point $P(x,y)$ on the circle, $\angle APB = 90°$ — the classic "angle in a semicircle" fact — so the chords $PA$ and $PB$ are perpendicular. Writing that perpendicularity in coordinate form gives

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

where: $(x_1,y_1)$ and $(x_2,y_2)$ are the diameter's endpoints. This form is a shortcut precisely when a problem gives you a diameter rather than a centre-and-radius pair.

General form. Expanding the standard form and relabelling the constants produces the general second-degree equation of a circle, always with equal $x^2$ and $y^2$ coefficients and no $xy$ term:

x^2 + y^2 + 2gx + 2fy + c = 0, \qquad \text{centre } (-g,-f), \ \text{radius } = \sqrt{g^2+f^2-c}

where: $g$ , $f$ , $c$ are the coefficients you read straight off a given equation once it's arranged in this exact layout. This is the form that makes three-point problems solvable by simultaneous equations, shown below.

Parametric form. Every point on a circle of centre $(h,k)$ and radius $r$ can be swept out by a single angle $\theta$ :

x = h + r\cos\theta, \qquad y = k + r\sin\theta, \qquad 0 \le \theta < 360°

where: $\theta$ is the angle measured from the centre. As $\theta$ runs through a full turn, $(x,y)$ traces the whole circle exactly once — useful for describing motion around a circle rather than just its static shape. The figure below shows how these forms describe the same circle from different starting data.

Figure

A circle with its centre marked, its radius drawn to a point on the rim, and a diameter drawn between two labelled endpoints, showing the data each form of the equation starts from

Derivation ACircle through three points

The goal is to build the full equation of a circle from three given points on it, using nothing beyond substitution into the general form and solving three simultaneous linear equations, so that a purely geometric fact — three non-collinear points determine exactly one circle — becomes a mechanical algebra problem with no guesswork.

Step 1. Start from the general form $x^2+y^2+2gx+2fy+c=0$ , which has exactly three unknowns, $g$ , $f$ , $c$ . Since a circle passes through each given point, substituting that point's coordinates for $(x,y)$ must make the equation true — so each point contributes one linear equation in $g$ , $f$ , $c$ .

Step 2. Substitute each of the three given points in turn to get three simultaneous linear equations. Because the unknowns $g$ , $f$ , $c$ appear only to the first power (the $x^2+y^2$ part is already a known number once a point is plugged in), this is genuinely a linear system, solvable by ordinary elimination or substitution — no quadratics to untangle.

Step 3. Solve the three equations together for $g$ , $f$ , $c$ . Whichever point makes the arithmetic simplest (often one lying on an axis, where a coordinate is zero) is worth substituting first, since it usually pins down one unknown immediately and shrinks the remaining system to two equations in two unknowns.

Step 4. Once $g$ , $f$ , $c$ are known, write the equation $x^2+y^2+2gx+2fy+c=0$ directly, and if a standard-form or centre/radius answer is wanted, read off centre $(-g,-f)$ and radius $\sqrt{g^2+f^2-c}$ . It is worth checking the result by substituting one of the original points back in — if the arithmetic doesn't return zero, an earlier step has a sign error.

Worked example

Find the equation of the circle passing through the points $A(0,0)$ , $B(4,0)$ , and $C(0,6)$ .

Step 1. Use the general form $x^2+y^2+2gx+2fy+c=0$ and substitute $A(0,0)$ : $0+0+0+0+c=0$ , so $c=0$ immediately, since every term except $c$ vanishes at the origin.

Step 2. Substitute $B(4,0)$ : $16+0+8g+0+c=0$ . With $c=0$ already known, this gives $16+8g=0$ , so $g=-2$ .

Step 3. Substitute $C(0,6)$ : $0+36+0+12f+c=0$ . With $c=0$ , this gives $36+12f=0$ , so $f=-3$ .

Step 4. Assemble the equation: $x^2+y^2-4x-6y=0$ . Centre $=(-g,-f)=(2,3)$ , radius $=\sqrt{g^2+f^2-c}=\sqrt{4+9-0}=\sqrt{13}$ . Check with $B(4,0)$ : distance from centre $(2,3)$ is $\sqrt{(4-2)^2+(0-3)^2}=\sqrt{4+9}=\sqrt{13}$ ✓, matching the radius found, so the equation is confirmed.

The circle through $A(0,0)$ , $B(4,0)$ , $C(0,6)$ is $x^2+y^2-4x-6y=0$ , with centre $(2,3)$ and radius $\sqrt{13}$ .

Competitive stretch

Does the circle $x^2+y^2-6x+8y-11=0$ pass through the same centre region, and what is its radius? Comparing with $x^2+y^2+2gx+2fy+c=0$ gives $2g=-6$ so $g=-3$ , $2f=8$ so $f=4$ , and $c=-11$ . Centre $=(-g,-f)=(3,-4)$ , radius $=\sqrt{g^2+f^2-c}=\sqrt{9+16+11}=\sqrt{36}=6$ . Completing the square directly on the original equation gives the same answer as a check: $x^2-6x+9+y^2+8y+16=11+9+16=36$ , so $(x-3)^2+(y+4)^2=36$ ✓. This second example is deliberately unrelated to the first — its point is to drill the coefficient-matching habit on an equation that is already expanded, which is the more common exam presentation than being handed three raw points.

Radius $6$ , centre $(3,-4)$ , standard form $(x-3)^2+(y+4)^2=36$ .

Common traps

Sign errors reading off the centre. The centre is $(-g,-f)$ , not $(g,f)$ — easy to slip on when the coefficients $2g$ and $2f$ are themselves negative in the given equation.
Forgetting the factor of 2. Matching $x^2+y^2-6x+8y-11=0$ against $x^2+y^2+2gx+2fy+c=0$ gives $2g=-6$ , hence $g=-3$ , not $g=-6$ .
Assuming any second-degree equation is a circle. It represents a genuine circle only if the $x^2$ and $y^2$ coefficients are equal, there is no $xy$ term, and $g^2+f^2-c>0$ ; otherwise it's a single point or has no real locus.
Reaching for the diameter-form condition without a diameter. That form encodes perpendicularity between two given points and any point on the circle — it only applies when the two given points are genuinely the ends of a diameter, not just any two points on the circle.

Ask the tutor: if solving the three simultaneous equations from a three-point circle problem feels error-prone, ask the AI tutor to walk through the elimination steps on a specific set of points.

Why does substituting three points into the general form give linear equations, when the general form itself has x squared and y squared terms?

Because $x$ and $y$ are known numbers once you plug in a specific point — only $g$ , $f$ , $c$ remain as unknowns, and they appear only to the first power in the equation. The $x^2+y^2$ part just becomes a fixed number added to the constant term. So each point really contributes one equation of the form $2gx_i+2fy_i+c=(\text{known number})$ , which is linear in $g$ , $f$ , $c$ — that's exactly why the three-point method reduces to solving a simultaneous linear system rather than a quadratic one.

How do I know whether to use the standard form, diameter form, or general form for a given problem?

Let the given data choose the form. If you're handed a centre and a radius (or enough to find them, like a centre and one point on the circle), the standard form is fastest. If you're given two points that are explicitly the ends of a diameter, the diameter form turns the perpendicularity fact directly into an equation with no extra work. If you're given three points on the circle, or an equation already in expanded polynomial form, the general form is the natural home, since it's built exactly for reading off or solving for $g$ , $f$ , $c$ .

If I expand the diameter form, do I get the same general form as usual?

Yes — expanding $(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$ gives $x^2+y^2-(x_1+x_2)x-(y_1+y_2)y+(x_1x_2+y_1y_2)=0$ , which matches $x^2+y^2+2gx+2fy+c=0$ with $2g=-(x_1+x_2)$ , $2f=-(y_1+y_2)$ , and $c=x_1x_2+y_1y_2$ . So the diameter form isn't a separate kind of circle equation — it's the general form with its coefficients pre-packaged in terms of the two diameter endpoints, which is exactly why it's faster when a diameter is given.

Equation of a circle

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads