1.03bPureCore

Circles

Q: What if completing the square leaves a negative on the right?

Then no real circle exists, since r^2 cannot be negative. Some questions are set precisely to elicit that conclusion, so state it rather than assuming an arithmetic slip.

A circle is the set of points a fixed distance from a fixed centre — which means its equation is really the distance formula with the square roots removed. OCR expects you to move between the two standard forms, and to use the geometric properties that make a tangent or chord question solvable.

35 min Coordinate Geometry

The big picture

Circle questions are where several strands of the course are deliberately brought together. Finding the centre from an expanded equation needs completing the square from the algebra chapter. Finding a tangent needs the perpendicular gradient condition from straight lines. Finding where a line cuts a circle needs simultaneous equations and the discriminant. That is why circles appear so often as a long, multi-part question: the examiner is testing whether the earlier tools have become usable rather than merely learned. Approach a circle question by asking which of those tools the geometry is pointing at.

What you'll be able to do

Write and interpret the equation $(x - a)^{2} + (y - b)^{2} = r^{2}$
Recover the centre and radius from an expanded equation by completing the square
Use the circle theorems for tangents, chords and angles in a semicircle
Find the intersection of a line and a circle, and interpret the discriminant

The equation is Pythagoras in disguise

A point $(x, y)$ lies on the circle of centre $(a, b)$ and radius $r$ exactly when its distance from the centre is $r$ . Writing that with the distance formula and squaring both sides removes the root and gives the standard form.

Read the form carefully: the numbers inside the brackets are the $negatives$ of the centre coordinates, and the right-hand side is $r^{2}$ , not $r$ . Both are routinely misread.

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

Centre

(a, b)

, radius

r

(x + 3)^{2}

(x - (- 3))^{2}

, so

a = - 3

(y - 1)^{2}

gives

b = 1

3The right side is

r^{2} = 25

, so

r = 5

AnswerCentre

(- 3, 1)

, radius

5

Tip — A plus inside the bracket means a negative coordinate. $(x + 3)^{2}$ places the centre at $x = - 3$ .

From the expanded form back to the centre

Exam questions usually give the equation expanded: $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ . To recover the centre and radius, complete the square in $x$ and in $y$ separately, then collect the constants.

This is the same completing-the-square procedure from the algebra chapter, done twice. It is the single most-tested skill in the whole circle topic.

1Group:

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

2Complete the square in

x

x^{2} - 6 x = (x - 3)^{2} - 9

3And in

y

y^{2} + 4 y = (y + 2)^{2} - 4

4Substitute:

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

, so

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

AnswerCentre

(3, - 2)

, radius

5

Why this works

If the constant ends up negative after collecting — say the right-hand side comes out as $- 4$ — there is no real circle, because a squared distance cannot be negative. That is a legitimate answer to a "show that no circle exists" question, not an error.

The three properties that solve circle questions

Almost every circle question turns on one of three facts, and each converts the problem into straight-line work you can already do.

$A tangent is perpendicular to the radius$ at the point of contact. So the tangent gradient is the negative reciprocal of the radius gradient — compute the radius gradient from the centre and the point, then flip and negate.

$The perpendicular from the centre bisects a chord$ . This gives right-angled triangles linking the radius, half the chord and the distance from centre to chord.

$The angle in a semicircle is a right angle$ . If three points on a circle form a right angle, the side opposite it is a diameter — which locates the centre as its midpoint.

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

Tangent ⟂ radius at the point of contact.

r^{2} = d^{2} + (\frac{chord}{2})^{2}

Radius, distance to chord, half-chord.

1Radius gradient from centre to

P

\frac{3 - ( - 1 )}{5 - 2} = \frac{4}{3}

2Tangent gradient is the negative reciprocal:

- \frac{3}{4}

3Point-gradient form at

P

y - 3 = - \frac{3}{4} (x - 5)

4Clear the fraction:

4 y - 12 = - 3 x + 15

Answer

3 x + 4 y - 27 = 0

Tip — Sketch the circle, the centre and the given point before starting. Which of the three properties applies is almost always obvious from a rough picture and rarely obvious from the algebra.

Lines meeting circles, and what the discriminant says

To find where a line meets a circle, substitute the line equation into the circle equation. The result is a quadratic, and its discriminant classifies the geometry.

Two real roots means the line cuts the circle twice. One repeated root means it touches — it is a tangent. No real roots means it misses entirely. This is the cleanest link in the course between an algebraic quantity and a geometric picture.

b^{2} - 4 a c > 0

Two intersections — a chord.

b^{2} - 4 a c = 0

One point of contact — a tangent.

b^{2} - 4 a c < 0

No intersection.

1Substitute the line into the circle:

x^{2} + (x + 2)^{2} = 2

2Expand and collect:

2 x^{2} + 4 x + 4 = 2

, so

2 x^{2} + 4 x + 2 = 0

, i.e.

x^{2} + 2 x + 1 = 0

3Discriminant:

2^{2} - 4 (1) (1) = 0

, so there is a repeated root and the line is a tangent.

4Solve

(x + 1)^{2} = 0

to get

x = - 1

, then

y = - 1 + 2 = 1

AnswerDiscriminant

0

, so it is a tangent; the point of contact is

(- 1, 1)

Why this works

Two answers for $k$ is exactly right geometrically: a line of fixed gradient can touch a circle on either side, so there are two parallel tangents.

Think like an examiner

Show the completing-the-square working in full. The centre and radius alone score poorly if the method is not visible.
State the radius, not $r^{2}$ . Leaving the answer as 25 when the radius is 5 is a common lost mark.
For tangent questions, state explicitly that the tangent is perpendicular to the radius — the reasoning is credited.
A discriminant argument needs a conclusion in words: " $b^{2} - 4 a c = 0$ , so the line is a tangent."

Common misconceptions

Reading

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

as centre

(4, 2)

, radius 9.

Centre

(- 4, 2)

and radius 3. Negate the bracket constants and square-root the right side.

Forgetting to subtract the completed-square constants.

x^{2} - 6 x

becomes

(x - 3)^{2} - 9

; the

- 9

must be carried across.

Using the radius gradient as the tangent gradient.

They are perpendicular. Take the negative reciprocal before writing the equation.

Assuming a line is a tangent because the question implies it.

Test it. The discriminant decides, and a positive value means a chord however the question is phrased.

Circle facts

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

Centre

(a, b)

, radius

r

x^{2} + y^{2} + 2 g x + 2 f y + c = 0

Expanded form — complete the square.

m_{tan} m_{rad} = - 1

Tangent ⟂ radius.

b^{2} - 4 a c = 0 \Rightarrow tangent

Discriminant classifies intersections.

Stretch yourself

A circle passes through $A (1, 1)$ , $B (7, 1)$ and $C (7, 9)$ . Show that $A C$ is a diameter, and hence find the equation of the circle.

Hint — Check the gradients of $A B$ and $B C$ . What does a right angle at $B$ tell you about $A C$ ?

Questions students ask

Why is the right-hand side

r^{2}

rather than

r

Because the equation comes from the distance formula with both sides squared, which removes the square root. Squaring is what makes the equation polynomial and workable.

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

Substitute it into the left-hand side of the standard form. A value less than

r^{2}

means inside, equal means on the circle, greater means outside.

What if completing the square leaves a negative on the right?

Then no real circle exists, since

r^{2}

cannot be negative. Some questions are set precisely to elicit that conclusion, so state it rather than assuming an arithmetic slip.

Is there a formula for the tangent I can just quote?

There are such formulae, but OCR expects the reasoning: radius gradient, negative reciprocal, point-gradient form. Working it through also survives variations a memorised formula does not cover.

Key takeaways

The circle equation is the distance formula squared: $(x - a)^{2} + (y - b)^{2} = r^{2}$ .
Complete the square in $x$ and $y$ to recover the centre and radius from expanded form.
A tangent is perpendicular to the radius at the point of contact.
The perpendicular from the centre bisects a chord; the angle in a semicircle is a right angle.
Substituting a line into a circle gives a quadratic whose discriminant classifies the intersection.

How this fits the course

Build on

Parametric Equations

Leads to

Parametric Equations

Test yourself

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Next step

Practise this topic

Real past-paper questions on Circles, marked mark-by-mark. How you do feeds straight into your weak-topic list, so your revision keeps targeting what actually needs work.

Practise this topic

Circles

The equation is Pythagoras in disguise

From the expanded form back to the centre

The three properties that solve circle questions

Lines meeting circles, and what the discriminant says

Practise this topic

Related topics

Circles

The equation is Pythagoras in disguise

From the expanded form back to the centre

The three properties that solve circle questions

Lines meeting circles, and what the discriminant says

Practise this topic

Related topics