OxfordAQA International AS and A-level Mathematics: Equation of a circle, tangents and circle geometry (9660)

This guide covers the circle for the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2. It follows content area PP1.1 (Circle) in section 3.2.1 of the specification. The specification places this content in International AS, Unit PSM1, and gives no further tier labels. Two companion pages are the revision notes and the practice questions. The course page is the OxfordAQA A-level Mathematics hub and the printable checklist lists every topic. If you are unsure where you stand, take a free 10-minute diagnostic first.

What PP1.1 requires

Content	Skills needed
Equation (x – a)² + (y – b)² = r²	Write it down from a centre and radius, and read both from it
Completing the square	Turn an expanded equation into a centre and a radius
Translation of circles	Say how the equation changes under a translation
Coordinate geometry of the circle	Use the angle in a semicircle, the line from the centre that meets a chord at right angles, and the tangent–radius right angle
Tangent and normal at a point	Find both equations using gradients. Implicit differentiation is not required

You also need straight-line coordinate geometry: gradient, midpoint, distance, perpendicular gradients (product −1) and the form y − y₁ = m(x − x₁). If any of that is shaky, revise it first.

1. The equation of a circle

A circle is every point at a fixed distance r (the radius) from a fixed point (a, b) (the centre). Applying the distance formula to a general point (x, y) and squaring gives

(x − a)² + (y − b)² = r²

Each bracket carries the opposite sign to the matching centre coordinate. So (x + 5)² + (y − 2)² = 16 has centre (−5, 2) and radius 4, because x + 5 = x − (−5).

Worked example 1. State the equation of the circle with centre (−5, 2) and radius 4, then expand it.

(x + 5)² + (y − 2)² = 16
x² + 10x + 25 + y² − 4y + 4 = 16
x² + y² + 10x − 4y + 13 = 0

Expanded equations always have x² and y² with equal coefficients and no xy term. That is how you recognise a possible circle.

Testing a point. Substitute (x, y) into (x − a)² + (y − b)². Equal to r² means the point is on the circle, less than r² means inside, and more than r² means outside.

2. Completing the square

To find the centre and radius from an expanded equation:

1. If x² and y² have a coefficient other than 1, divide every term by it.
2. Group the x terms and the y terms.
3. Rewrite each variable using x² + px = (x + p/2)² − p²/4, and the same for y.
4. Move the constants to the right-hand side. That number is r².

Worked example 2. Find the centre and radius of x² + y² + 8x − 6y − 24 = 0.

(x + 4)² − 16 + (y − 3)² − 9 − 24 = 0
(x + 4)² + (y − 3)² = 49

Centre (−4, 3), radius 7.

Worked example 3. Find the centre and radius of 2x² + 2y² + 12x − 8y − 10 = 0.

x² + y² + 6x − 4y − 5 = 0          (divide by 2)
(x + 3)² − 9 + (y − 2)² − 4 − 5 = 0
(x + 3)² + (y − 2)² = 18

Centre (−3, 2), radius √18 = 3√2. Leave the radius as a surd unless the question asks for a decimal.

When it is not a circle. If the right-hand side comes out zero, the “circle” is a single point. If it is negative, no points satisfy the equation. For example x² + y² + 2x − 6y + 15 = 0 gives (x + 1)² + (y − 3)² = −5, so there is no circle.

3. Translation of circles

A translation moves every point without turning or resizing it. So the radius is unchanged and only the centre moves.

If the circle (x − a)² + (y − b)² = r² is translated by the vector (p, q), the centre moves from (a, b) to (a + p, b + q). The image is

(x − a − p)² + (y − b − q)² = r²

Each x is replaced by (x − p) and each y by (y − q).

Worked example 4. The circle (x − 1)² + (y + 2)² = 9 is translated by the vector (3, 4). Find the equation of the image.

The centre (1, −2) moves to (1 + 3, −2 + 4) = (4, 2). The radius stays 3.

Image: (x − 4)² + (y − 2)² = 9, which expands to x² + y² − 8x − 4y + 11 = 0.

To find an unknown translation, compare the two centres: the vector is image centre minus original centre.

4. Circle properties in coordinate form

The specification names three properties. Each one turns into a gradient or distance calculation.

The angle in a semicircle is a right angle. Suppose AB is a diameter and P is any other point on the circle. Then AP and BP are perpendicular, so (gradient AP) × (gradient BP) = −1. The converse also holds: if the gradients of AP and BP multiply to −1, then P is on the circle with diameter AB.

Worked example 5. The segment from A(−2, 3) to B(6, 7) is a diameter. Work out the equation of the circle, and prove that P(4, 9) lies on it.

centre = midpoint of AB = (2, 5)
r² = (−2 − 2)² + (3 − 5)² = 16 + 4 = 20
equation: (x − 2)² + (y − 5)² = 20

Check P: (4 − 2)² + (9 − 5)² = 4 + 16 = 20, so P is on the circle.

Check with the semicircle property: gradient AP = (9 − 3)/(4 + 2) = 1, gradient BP = (9 − 7)/(4 − 6) = −1. The product is −1, so angle APB is a right angle. Both methods agree.

A perpendicular dropped from the centre onto a chord splits it into two equal halves. This gives a right-angled triangle: centre, midpoint of the chord, and one end of the chord. Then r² = d² + (half the chord)², where d measures how far the chord lies from the centre.

In reverse, every chord has a perpendicular bisector that goes through the centre. Two chords therefore fix the centre where their perpendicular bisectors cross.

Worked example 6. A circle has centre C(4, 1) and radius √50. A chord has midpoint M(1, 7). Find the length of the chord.

CM² = (1 − 4)² + (7 − 1)² = 9 + 36 = 45
half-chord² = 50 − 45 = 5
chord length = 2√5

Note that the midpoint is given, so CM is perpendicular to the chord automatically.

5. Tangents and normals

A tangent meets the circle at exactly one point and is at right angles to the radius there. The normal at a point is the line perpendicular to the tangent there, so it runs along the radius and passes through the centre.

No calculus is needed. The method is:

1. Find the centre C and the point of contact P.
2. Gradient of radius CP = (y_P − y_C)/(x_P − x_C).
3. Gradient of tangent = −1 ÷ (gradient of CP). Gradient of normal = gradient of CP.
4. Use y − y₁ = m(x − x₁) with P as (x₁, y₁).

If the radius is horizontal, the tangent is vertical (x = constant). If the radius is vertical, the tangent is horizontal (y = constant). Do not divide by zero.

Worked example 7. Find the tangent and normal to (x − 3)² + (y + 1)² = 40 at P(5, 5).

Check P is on the circle: (5 − 3)² + (5 + 1)² = 4 + 36 = 40. ✓

gradient of CP = (5 − (−1))/(5 − 3) = 6/2 = 3
tangent gradient = −1/3
tangent:  y − 5 = −(1/3)(x − 5)   →   x + 3y = 20
normal:   y − 5 = 3(x − 5)        →   y = 3x − 10

Check: the normal should pass through the centre (3, −1). 3(3) − 10 = −1. ✓

Length of a tangent from an outside point. The tangent, the radius to the contact point and the line from the outside point to the centre form a right-angled triangle, with the right angle at the contact point. So

(tangent length)² = (distance from point to centre)² − r²

Worked example 8. Find the length of a tangent drawn from T(9, 7) to the circle in example 7.

TC² = (9 − 3)² + (7 + 1)² = 36 + 64 = 100
tangent length² = 100 − 40 = 60
tangent length = 2√15

Common errors

• Reading the centre with the wrong signs: (x + 3)² means a centre at x = −3, not +3.
• Forgetting to divide by the coefficient of x² before completing the square.
• Giving r² as the radius. After completing the square the right-hand side is r², so take the square root.
• Using the radius gradient as the tangent gradient. The tangent gradient is its negative reciprocal.
• Skipping the check that the given point lies on the circle before finding the tangent.
• In a translation, moving the centre the wrong way. A translation by (3, 4) adds 3 to x and 4 to y.
• Finding the equation of the tangent in a form that is not simplified when the question asks for whole-number coefficients.

Working by hand

Every example here is short algebra that you can do by hand, so write each step out. A clear line of working shows your method even if you slip on the arithmetic later.

Next steps

Work through the practice questions, then use the revision notes for quick recall. The wider Unit P1 pure maths guide covers the straight-line and quadratics skills used here. The diagnostics show which topics to prioritise.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660), Version 5.2, section 3.2.1 PP1.1 (Circle), published by OxfordAQA. For International AS exams from May/June 2018 and International A-level exams from May/June 2019.