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

Condensed notes for OxfordAQA International AS and A-level Mathematics (9660), Version 5.2, content area PP1.1 (Circle). The specification places this in International AS, Unit PSM1. For full explanations read the study guide, then test yourself with the practice questions. Course pages: hub and checklist. To find weak spots, try a free 10-minute diagnostic.

Key results

Item	Result
Circle, centre (a, b), radius r	(x − a)² + (y − b)² = r²
Expanded form	x² + y² + px + qy + c = 0 (equal x² and y² coefficients, no xy term)
Centre from expanded form	(−p/2, −q/2)
Radius from expanded form	r² = p²/4 + q²/4 − c
Translation by (s, t)	Centre (a, b) → (a + s, b + t); radius unchanged
Gradient of tangent	−1 ÷ gradient of the radius at that point
Gradient of normal	Same as the radius; the normal passes through the centre
Tangent length from outside point T	√(TC² − r²)

The centre and radius formulas follow from completing the square, so you can rebuild them in the exam instead of memorising them.

Method: centre and radius from an expanded equation

1. Divide through so the x² and y² coefficients are 1.
2. Rewrite the x terms and then the y terms as squares.
3. Move constants right. The right side is r².
4. Check r² > 0. Zero means a single point; negative means no circle.

Reminder. x² + y² − 12x + 6y + 9 = 0:

(x − 6)² − 36 + (y + 3)² − 9 + 9 = 0
(x − 6)² + (y + 3)² = 36

Centre (6, −3), radius 6.

Method: inside, on or outside

Compute (x − a)² + (y − b)² for the point and compare with r².

Result	Position
Less than r²	Inside
Equal to r²	On the circle
Greater than r²	Outside

Method: translating a circle

1. Write the original centre and radius.
2. Add the vector to the centre.
3. Keep the radius. Write the new equation.

Reminder. (x + 3)² + (y − 1)² = 49 translated by (2, −5): centre (−3, 1) → (−1, −4). Image: (x + 1)² + (y + 4)² = 49.

To find an unknown vector: image centre − original centre.

The three circle properties

Property	What to calculate
Angle in a semicircle is 90°	Gradient AP × gradient BP = −1 when AB is a diameter
Perpendicular dropped from the centre onto a chord	It splits the chord into equal halves: r² = d² + (half-chord)², with d the centre-to-chord distance
Tangent ⟂ radius at the point of contact	Gradient of tangent × gradient of radius = −1

Uses of the chord property:

• The centre lies on the perpendicular bisector of every chord.
• Two chords give the centre as the intersection of their perpendicular bisectors.
• A line through the centre perpendicular to a chord meets it at its midpoint.

Method: tangent and normal at a point

1. Check the point lies on the circle.
2. Gradient of radius = (change in y)/(change in x) from the centre to the point.
3. Tangent gradient = negative reciprocal. Normal gradient = radius gradient.
4. Use y − y₁ = m(x − x₁).
5. Tidy to the form asked for, such as ax + by = c with integers.
6. Check the normal passes through the centre.

Special cases. Horizontal radius gives vertical tangent x = constant. Vertical radius gives horizontal tangent y = constant.

Reminder. Circle (x − 2)² + (y − 3)² = 52 at (8, 7): 36 + 16 = 52 ✓. Radius gradient = 4/6 = 2/3. Tangent gradient −3/2: y − 7 = −(3/2)(x − 8), so 3x + 2y = 38. Normal: y − 7 = (2/3)(x − 8), so 2x − 3y + 5 = 0. Centre check: 4 − 9 + 5 = 0 ✓.

Implicit differentiation is not required by the specification. Gradients come from the centre and the point of contact.

More worked reminders

Semicircle check. A(0, 1) and B(6, 5) are the ends of a diameter. Is P(5, 6) on the circle? Gradient AP = (6 − 1)/5 = 1, gradient BP = (6 − 5)/(5 − 6) = −1. Product −1, so angle APB is 90° and P is on the circle. Direct check: centre (3, 3), r² = 13, and (5 − 3)² + (6 − 3)² = 13 ✓.

Finding a centre from a chord. A circle passes through (1, 0) and (7, 4), and its centre is on the y-axis. The chord has midpoint (4, 2) and gradient 4/6 = 2/3, so the perpendicular bisector has gradient −3/2: y − 2 = −(3/2)(x − 4). At x = 0, y = 8. Centre (0, 8), r² = 1 + 64 = 65. Check the other point: 49 + 16 = 65 ✓.

Chord length from a distance. Radius 13, chord 5 from the centre. Half-chord² = 169 − 25 = 144, half-chord 12, chord 24.

Tangent length. Circle (x − 1)² + (y − 1)² = 9, external point (7, 9). TC² = 36 + 64 = 100. Tangent length² = 100 − 9 = 91, so the length is √91.

Line meeting a circle. Substitute the line into the circle equation and get a quadratic. Two roots mean two intersections (a chord), a repeated root means the line is a tangent, and no real roots mean no intersection. This uses quadratics knowledge from the Unit P1 content, so see the Pure Maths 1 revision notes if it feels rusty.

Exam-style patterns to recognise

Wording in the question	What it is asking for
“Find the centre and radius”	Complete the square on x and y
“AB is a diameter”	Midpoint is the centre; half the distance AB is the radius
“Prove P is on the circle”	Put its coordinates into the left-hand side and obtain r²
“The tangent at P”	Perpendicular to the radius CP
“The normal at P”	The line through P and the centre
“The circle is translated by”	Move the centre only
“Touches the x-axis”	Radius equals the distance from the centre to the axis
“Touches the y-axis”	Radius equals the distance from the centre to that axis

Must-know distinctions

• Radius against r². The equation gives r². Take the square root.
• Tangent against normal. The tangent meets the radius at 90°; the normal runs along the radius.
• Centre sign flip. (x − a) gives centre a. (x + a) gives centre −a.
• Translation against enlargement. A translation keeps r. Only the centre moves.
• Chord against diameter. A diameter is a chord through the centre. Any chord’s perpendicular bisector passes through the centre.

Quick self-test

1. Give the equation of the circle centred at (−3, 4) with radius 7.
2. Find the centre and radius of x² + y² − 6x + 10y + 18 = 0.
3. Find the centre and radius of 4x² + 4y² − 8x + 16y − 20 = 0.
4. Is the point (6, −2) inside, on or outside the circle (x − 2)² + (y + 5)² = 30?
5. The circle (x − 4)² + (y − 3)² = 11 is translated by the vector (−6, 2). State the image equation.
6. A diameter has endpoints (0, 1) and (6, 5). Write the equation of the circle.
7. A chord of a circle with radius 10 lies 6 units from the centre. How long is the chord?
8. Find the equation of the tangent to the circle x² + y² = 29 at (5, 2).
9. Find the equation of the normal to (x − 1)² + (y − 1)² = 25 at (4, 5).
10. Find the length of the tangent from (0, 8) to the circle x² + y² = 20.
11. Is x² + y² − 4x + 2y + 9 = 0 a circle? Give a reason.

Answers

1.  (x + 3)² + (y − 4)² = 49
2.  (x − 3)² + (y + 5)² = 9 + 25 − 18 = 16 → centre (3, −5), radius 4
3.  ÷4: x² + y² − 2x + 4y − 5 = 0 → (x − 1)² + (y + 2)² = 10
    centre (1, −2), radius √10
4.  (6 − 2)² + (−2 + 5)² = 16 + 9 = 25 < 30 → inside
5.  centre (4, 3) → (−2, 5): (x + 2)² + (y − 5)² = 11
6.  centre (3, 3); r² = 9 + 4 = 13 → (x − 3)² + (y − 3)² = 13
7.  half-chord² = 100 − 36 = 64 → half-chord 8 → chord 16
8.  radius gradient 2/5, tangent gradient −5/2:
    y − 2 = −(5/2)(x − 5) → 5x + 2y = 29
9.  radius gradient (5 − 1)/(4 − 1) = 4/3; normal passes through the
    point with that gradient: y − 5 = (4/3)(x − 4) → 4x − 3y = 1
10. TC² = 64, tangent² = 64 − 20 = 44 → 2√11
11. (x − 2)² + (y + 1)² = 4 + 1 − 9 = −4 < 0 → no, no real points

Question 9 check: (4 − 1)² + (5 − 1)² = 25 ✓ (the point is on the circle), and the centre (1, 1) gives 4 − 3 = 1 ✓.

Where marks are usually lost

• Writing the centre with the wrong sign after completing the square.
• Stating r² as the radius.
• Not dividing by the common coefficient first, so the completed squares are wrong.
• Using the radius gradient as the tangent gradient, or flipping without negating.
• Starting the tangent equation without checking the point is on the circle, which can hide a copying error.
• Moving the centre in the wrong direction in a translation.
• Showing a bare answer for a “show that” question instead of every line of substitution.
• Leaving a tangent equation untidy when integer coefficients are requested.
• Rounding a surd radius early. Keep exact values until the end.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660), Version 5.2, section 3.2.1 PP1.1 (Circle), published by OxfordAQA.