C2PureCore

Equation of a Circle

Q: Where does the circle equation come from?

From the distance formula. A point (x,y) is on the circle when its distance from the centre (a,b) equals the radius r. Writing that with Pythagoras and squaring gives (x-a)^2 + (y-b)^2 = r^2.

Q: How do I find the centre and radius from an expanded equation?

Complete the square separately in x and in y, then rearrange into (x-a)^2 + (y-b)^2 = r^2. The centre is (a,b) and the radius is r^2.

Q: Why is the centre’s sign the opposite of what’s in the bracket?

Because the bracket is (x - a). To read a, ask what value makes the bracket zero: for (x + 4) that is x = -4, so the centre’s x-coordinate is -4.

A circle is every point a fixed distance from a centre — and writing that sentence with Pythagoras gives its equation directly. A few geometric facts (tangent perpendicular to radius, angle in a semicircle) then unlock almost every circle question.

30 min Video by TLMaths Coordinate Geometry

The big picture

The circle is the first curve you meet whose equation is not a function (it fails the vertical line test), and understanding it cements the idea that an equation describes a $set of points$ rather than an input–output rule. The elegant part is that its equation is nothing new: it is the distance formula in disguise. Combine that with a handful of classical circle theorems — the tangent meeting the radius at a right angle, the perpendicular bisector of a chord passing through the centre — and you can find tangents, centres and radii from surprisingly little information. It also gives completing the square a real job to do.

A-Level Maths C2-01: Introducing the Equation of the CircleWatch the walkthrough, then work through the notes below.

Open on YouTube

What you'll be able to do

Write and read the equation $(x - a)^{2} + (y - b)^{2} = r^{2}$
Find the centre and radius from an expanded circle equation by completing the square
Use the tangent–radius right angle to find tangent lines
Apply key circle facts (semicircle angle, chord bisector)

The equation is the distance formula

A circle is the set of all points a fixed distance $r$ (the radius) from a fixed point $(a, b)$ (the centre). Take any point $(x, y)$ on it: the distance from the centre must be $r$ . Apply Pythagoras to the horizontal and vertical gaps and square both sides — and out drops the circle equation.

So there is nothing to memorise separately: the circle equation is just “distance from the centre equals $r$ ”, written with the distance formula.

(x - a)^{2} + (y - b)^{2} = r^{2}

Centre

(a, b)

, radius

r

. Note the signs flip: centre is

(+ a, + b)

1Compare with

(x - a)^{2} + (y - b)^{2} = r^{2}

x - 3

gives

a = 3

;

y + 2 = y - (- 2)

gives

b = - 2

r^{2} = 25

, so

r = 5

AnswerCentre

(3, - 2)

, radius

5

Tip — Watch the signs: $(y + 2)^{2}$ means $b = - 2$ , not $+ 2$ . The centre coordinates are the values that make each bracket zero.

Finding the centre from the expanded form

A circle equation is often given expanded, like $x^{2} + y^{2} - 6 x + 4 y - 12 = 0$ . To recover the centre and radius, $complete the square$ in $x$ and in $y$ separately — turning the expanded mess back into centre–radius form.

1Group:

(x^{2} - 6 x) + (y^{2} + 4 y) = 12

2Complete each square:

(x - 3)^{2} - 9 + (y + 2)^{2} - 4 = 12

3Tidy:

(x - 3)^{2} + (y + 2)^{2} = 25

AnswerCentre

(3, - 2)

, radius

5

Why this works

This is completing the square earning its keep. The $- 9$ and $- 4$ that appear are the “corrections” from the two squares — move them to the right-hand side and the constant that remains is $r^{2}$ .

Tangents and the right-angle fact

The single most useful circle fact: a $\textbf{tangent meets the radius at the point of contact at$ 90° $}$ . So to find a tangent, find the gradient of the radius to that point, then take its negative reciprocal for the tangent gradient — the straight-line skills from the last lesson do the rest.

Two more facts appear often. The angle in a $semicircle$ (subtended by a diameter) is a right angle. And the $perpendicular from the centre to a chord bisects it$ — so the perpendicular bisector of any chord passes through the centre.

m_{radius} \times m_{tangent} = - 1

Tangent ⟂ radius at the point of contact.

1Gradient of the radius from

(1, 2)

(4, 6)

\frac{6 - 2}{4 - 1} = \frac{4}{3}

2Tangent is perpendicular: negative reciprocal of

\frac{4}{3}

- \frac{3}{4}

AnswerTangent gradient

- \frac{3}{4}

Why this works

Almost every hard circle question is really a straight-line question in disguise: use a circle fact to get a right angle or a midpoint, then finish with gradients, midpoints and equations of lines.

Think like an examiner

Read the centre with the correct signs: $(x + 5)^{2}$ means the $x$ -coordinate of the centre is $- 5$ .
Do not confuse $r$ and $r^{2}$ — the right-hand side of the equation is the radius $squared$ .
For tangents, state the tangent–radius right angle explicitly; it justifies using the negative reciprocal gradient.
When given the expanded form, completing the square in both variables is the expected method for the centre and radius.

Common misconceptions

Reading the centre of

(x + 4)^{2} + (y - 1)^{2} = 9

(4, 1)

It is

(- 4, 1)

: the centre values make each bracket zero.

Giving the radius of

\dots = 16

as 16.

The right-hand side is

r^{2} = 16

, so

r = 4

Using the radius gradient itself as the tangent gradient.

The tangent is perpendicular to the radius — use the negative reciprocal.

Forgetting to move the square-completion corrections across when finding

r^{2}

The

- a^{2}

and

- b^{2}

terms belong on the right; only what remains there is

r^{2}

Circle essentials

(x - a)^{2} + (y - b)^{2} = r^{2}

Centre

(a, b)

, radius

r

m_{rad} m_{tan} = - 1

Tangent ⟂ radius.

angle in a semicircle = 9 0^{\circ}

Diameter subtends a right angle.

centre ⊥ chord bisects it

Chord bisector through centre.

Stretch yourself

A circle has equation $(x - 2)^{2} + (y - 3)^{2} = 25$ . Find the equation of the tangent to the circle at the point $(5, 7)$ .

Hint — Check the point is on the circle, find the radius gradient to it, take the negative reciprocal for the tangent gradient, then use point-gradient form.

Questions students ask

Where does the circle equation come from?

From the distance formula. A point

(x, y)

is on the circle when its distance from the centre

(a, b)

equals the radius

r

. Writing that with Pythagoras and squaring gives

(x - a)^{2} + (y - b)^{2} = r^{2}

How do I find the centre and radius from an expanded equation?

Complete the square separately in

x

and in

y

, then rearrange into

(x - a)^{2} + (y - b)^{2} = r^{2}

. The centre is

(a, b)

and the radius is

r^{2}

How do I find the equation of a tangent to a circle?

Use the fact that the tangent is perpendicular to the radius at the point of contact. Find the radius gradient, take its negative reciprocal for the tangent gradient, then use the point-gradient form of a line.

Why is the centre’s sign the opposite of what’s in the bracket?

Because the bracket is

(x - a)

. To read

a

, ask what value makes the bracket zero: for

(x + 4)

that is

x = - 4

, so the centre’s

x

-coordinate is

- 4

Key takeaways

A circle is all points distance $r$ from centre $(a, b)$ : $(x - a)^{2} + (y - b)^{2} = r^{2}$ .
Read centre coordinates as the values making each bracket zero; the RHS is $r^{2}$ .
Complete the square in $x$ and $y$ to find the centre/radius from expanded form.
Tangent ⟂ radius: use the negative reciprocal of the radius gradient.
Circle questions usually reduce to straight-line work after one geometric fact.

How this fits the course

Build on

The Discriminant (line meets circle)

Leads to

Parametric Equations

Test yourself

Ready to lock in Equation of a Circle? Pick a mode and earn XP & Dobloons.