Cambridge International AS & A Level Mathematics 9709: Coordinate Geometry (Pure Mathematics 1) – Revision Notes

These notes condense section 1.3, Coordinate geometry, of the Cambridge International AS & A Level Mathematics 9709 syllabus for exams in 2026 and 2027 (Version 4). It is part of Pure Mathematics 1 and is examined on Paper 1 (1 hour 50 minutes, 75 marks), which is compulsory for both AS Level and A Level. A scientific calculator is allowed, but unsupported calculator answers earn no marks, so show the algebra. For full explanations, use the coordinate geometry study guide.

Other links: coordinate geometry practice questions, the quadratics revision notes (the discriminant work used below), the Pure Mathematics 1 mixed practice, the A Level Mathematics hub, the printable 9709 checklist, the free AS Level diagnostic and the 9709 self-check question bank.

What 1.3 asks you to do

The syllabus lists five learning outcomes. You should be able to:

1. find the equation of a straight line given sufficient information (two points, or a point and the gradient);
2. interpret and use y = mx + c, y − y₁ = m(x − x₁) and ax + by + c = 0, including distances, gradients, midpoints, points of intersection and the gradients of parallel and perpendicular lines;
3. understand that (x − a)² + (y − b)² = r² is the circle with centre (a, b) and radius r, including the expanded form x² + y² + 2gx + 2fy + c = 0;
4. use algebraic methods for problems on lines and circles, using tangent perpendicular to radius, the angle in a semicircle and symmetry;
5. relate a graph to its equation, and use the link between points of intersection and solutions of equations (for example, the values of k for which y = x + k meets, touches or misses a curve).

Implicit differentiation is not included. Find tangents to circles with the radius, not with calculus.

Formulas

None of these is in the MF19 formulae list, so learn them all.

Result	Formula
Gradient of AB	m = (y₂ − y₁)/(x₂ − x₁)
Midpoint of AB	((x₁ + x₂)/2, (y₁ + y₂)/2)
Length of AB	√((x₂ − x₁)² + (y₂ − y₁)²)
Line through (x₁, y₁), gradient m	y − y₁ = m(x − x₁)
Parallel lines	m₁ = m₂
Perpendicular lines	m₁m₂ = −1, so m₂ = −1/m₁
Circle, centre (a, b), radius r	(x − a)² + (y − b)² = r²
Expanded circle	x² + y² + 2gx + 2fy + c = 0
Centre and radius of expanded form	centre (−g, −f), r = √(g² + f² − c)

Straight lines

Reading a line. Rearrange to y = mx + c to read the gradient and y-intercept. From ax + by + c = 0 the gradient is −a/b.

Which form to use.

• You have a point and a gradient: y − y₁ = m(x − x₁). It needs no rearranging first.
• You need the y-intercept: y = mx + c.
• The question says “in the form ax + by + c = 0” or “with integer coefficients”: clear fractions and collect everything on one side.

Method: perpendicular bisector of AB

1. Midpoint M of AB.
2. Gradient of AB, then the perpendicular gradient −1/m.
3. y − y_M = (−1/m)(x − x_M).

Worked reminder. A(−4, 1), B(2, 5). M = (−1, 3). Gradient AB = 4/6 = 2/3, so the perpendicular gradient is −3/2. y − 3 = −(3/2)(x + 1) gives 3x + 2y − 3 = 0.

Intersection of two lines. Solve the equations simultaneously. Elimination is usually cleaner when both are in ax + by + c = 0 form.

Foot of the perpendicular from P to a line. Write the line through P with the perpendicular gradient, then intersect it with the original line. The distance from P to that foot is the shortest distance from P to the line.

Area of a triangle. Look for a right angle (gradients multiplying to −1) or an isosceles shape (a vertex on a perpendicular bisector). Then area = ½ × base × perpendicular height, with both lengths from the distance formula.

Circles

Centre-radius form. (x − a)² + (y − b)² = r². Watch the signs: (x + 3)² + (y − 1)² = 49 has centre (−3, 1) and radius 7.

Method: expanded form to centre and radius

1. Group: (x² + 2gx) + (y² + 2fy) = −c.
2. Complete the square on x and on y separately.
3. Move the constants: (x + g)² + (y + f)² = g² + f² − c.
4. Centre (−g, −f); radius = √(g² + f² − c).

Worked reminder. x² + y² − 10x + 4y + 13 = 0 gives (x − 5)² − 25 + (y + 2)² − 4 + 13 = 0, so (x − 5)² + (y + 2)² = 16. Centre (5, −2), radius 4.

If g² + f² − c ≤ 0, the equation is not a real circle.

Inside, on or outside? Substitute the point into (x − a)² + (y − b)² and compare with r²: less means inside, equal means on the circle, greater means outside.

Finding a circle’s equation. You need the centre and the radius.

• Ends of a diameter given: centre = midpoint, radius = half the diameter.
• Centre given and a point on the circle: r² = (distance from centre to the point)².
• Two points and a line containing the centre: the centre is where the perpendicular bisector of the two points meets that line.
• Three points: the centre is where two perpendicular bisectors meet.

Geometry of the circle

Property	How it is used
Tangent perpendicular to radius	Tangent gradient = −1/(gradient of radius to the point of contact)
Angle in a semicircle is 90°	If AB is a diameter and C is on the circle, AC ⊥ BC (gradients multiply to −1)
Symmetry	The perpendicular bisector of any chord passes through the centre; the line from the centre to a chord’s midpoint is perpendicular to the chord

Method: tangent at a point P on the circle

1. Find the centre C.
2. Gradient of CP.
3. Tangent gradient = −1/(gradient of CP).
4. y − y_P = m(x − x_P).

If the question says “show that P lies on the circle”, substitute P and show the equation is satisfied before you use it.

Intersections, graphs and the discriminant

The points where two graphs meet are the solutions of their equations solved together.

Method: line and curve (or line and circle)

1. Make y (or x) the subject of the linear equation.
2. Substitute into the curve or circle to get one quadratic.
3. Solve; then find the other coordinate from the LINE.

Worked reminder. y = x − 3 with x² + y² − 10x + 4y + 13 = 0 gives 2x² − 12x + 10 = 0, so x² − 6x + 5 = 0, x = 1 or 5. Points (1, −2) and (5, 2).

The discriminant of that quadratic tells you how many times they meet:

b² − 4ac	Line and curve
> 0	meet at two distinct points
= 0	line is a tangent (touches once)
< 0	do not meet

For “the line meets the curve” (at least once) use b² − 4ac ≥ 0. When the line contains an unknown such as k or m, the discriminant becomes an inequality in that unknown. Solve it as a quadratic inequality if it contains k².

Must-know distinctions

• Parallel vs perpendicular. Parallel: same gradient. Perpendicular: product −1. A gradient of 2 is perpendicular to −1/2, not to −2 or 1/2.
• Midpoint vs perpendicular bisector. The midpoint is a point; the perpendicular bisector is a line through it.
• r vs r². In (x − a)² + (y − b)² = 20 the radius is √20 = 2√5, not 20.
• Centre signs. (x − a)² gives +a; x² + 2gx gives −g.
• Tangent vs radius. The tangent at P is perpendicular to CP; it does not pass through the centre.
• Two points vs at least one. “Two distinct points” is b² − 4ac > 0; “meets” is ≥ 0; “touches” is = 0.
• Distance vs distance squared. Compare squared distances with r² when testing a point; take the square root only for a final length.

Quick self-test

1. Find the gradient and the midpoint of the line joining (−2, 5) and (4, −7).
2. Find the distance between (−1, 3) and (4, −9).
3. State the gradient and y-intercept of 4x − 5y + 10 = 0.
4. Find the line through (2, −3) perpendicular to y = 4x − 1, in the form ax + by + c = 0.
5. Are 6x + 4y = 9 and y = −1.5x + 2 parallel, perpendicular or neither?
6. Find the centre and radius of x² + y² + 6x − 8y − 11 = 0.
7. Explain why x² + y² − 4x + 2y + 10 = 0 is not a circle.
8. Find the equation of the circle with centre (2, −5) passing through (−1, −1).
9. Is (7, 1) inside, on or outside the circle (x − 3)² + (y + 2)² = 30?
10. For y = x + k and y = x² − x + 3, find the value of k for which the line is a tangent, and the set of values for which they meet at two distinct points.
11. Find the tangent to x² + y² = 25 at (3, −4).
12. A diameter of a circle has ends (−3, 2) and (5, 8). Find the equation of the circle.

Answers

1. Gradient = −12/6 = −2; midpoint (1, −1).
2. √(5² + 12²) = 13.
3. y = (4/5)x + 2: gradient 4/5, y-intercept 2.
4. Gradient −1/4: y + 3 = −(1/4)(x − 2), so x + 4y + 10 = 0.
5. 6x + 4y = 9 gives y = −1.5x + 9/4. Same gradient, different intercept: parallel.
6. (x + 3)² + (y − 4)² = 9 + 16 + 11 = 36: centre (−3, 4), radius 6.
7. g² + f² − c = 4 + 1 − 10 = −5 < 0, so no real radius.
8. r² = 3² + 4² = 25: (x − 2)² + (y + 5)² = 25.
9. 4² + 3² = 25 < 30: inside.
10. x² − 2x + 3 − k = 0; discriminant 4k − 8. Tangent: k = 2. Two points: k > 2.
11. Radius gradient −4/3, so tangent gradient 3/4: y + 4 = (3/4)(x − 3), 3x − 4y − 25 = 0.
12. Centre (1, 5), r² = 4² + 3² = 25: (x − 1)² + (y − 5)² = 25.

Where marks are usually lost

• Using −a/b wrongly: the gradient of 5x − 2y + 1 = 0 is 5/2, not −5/2 or 5.
• Taking the negative gradient (−m) instead of the negative reciprocal (−1/m) for a perpendicular line.
• Writing the perpendicular bisector through A or B instead of through the midpoint.
• Stopping at y = mx + c with fractions when the question asks for ax + by + c = 0 with integer coefficients.
• Sign errors in the centre: (x + 2)² + (y − 7)² = 9 has centre (−2, 7).
• Giving r² as the radius, or leaving the radius as a decimal when an exact surd was expected.
• Finding the tangent with the gradient of the radius itself, forgetting the perpendicular step.
• Finding the second coordinate of an intersection from the circle, which gives extra (wrong) points; use the line.
• Using > 0 when the question says “meets” or “intersects”, which needs ≥ 0.
• Not showing that a given point lies on the circle before using it in a “show that” question.

Official syllabus

Cambridge International AS & A Level Mathematics 9709 syllabus, for exams in 2026 and 2027 (Version 4), Cambridge University Press & Assessment. Section 1 Pure Mathematics 1 (for Paper 1): 1.3 Coordinate geometry.