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A Level Maths 9709 · Paper 1 Pure Mathematics 1 · 1.3 · ~8 min read

Coordinate geometry: lines and circles

Equations of straight lines, parallel and perpendicular gradients, perpendicular bisectors, the equation of a circle in both forms, tangents, chords and where lines meet circles.

Gradient & line equationsParallel & perpendicularMidpoint & distancePerpendicular bisectorCircle equationCentre & radiusTangentsLine meets circle
RSby Riyaz Sir
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Paper tag = where it is examinedFORMULA LIST (MF19) given in the examM1A1B1 how marks are awardedhighlight = comes up again and againTest yourself at the end

In 30 seconds

  • A straight line needs a point and a gradient. Use y − y₁ = m(x − x₁): it is quicker and safer than y = mx + c.
  • Perpendicular gradients multiply to −1: if one gradient is m, the other is −1/m.
  • A circle with centre (a, b) and radius r is (x − a)² + (y − b)² = r². To find the centre and radius from the expanded form, complete the square in x and in y.
  • Three circle facts do most of the work: a tangent is perpendicular to the radius; the angle in a semicircle is 90°; the perpendicular bisector of any chord passes through the centre.

By the end you can

  • Find the equation of a line from two points, or a point and a gradient, in any required form.
  • Use parallel and perpendicular gradients, midpoints and distances.
  • Find the centre and radius of a circle and write its equation.
  • Find tangents to circles and decide whether a line cuts, touches or misses a circle.

1Straight lines

The toolkit

m = (y₂ − y₁)/(x₂ − x₁)
gradient through (x₁, y₁) and (x₂, y₂)
y − y₁ = m(x − x₁)
line through (x₁, y₁) with gradient m
  • Other forms: y = mx + c (gradient m, y-intercept c); ax + by + c = 0 (rearrange to read off m = −a/b).
  • Midpoint of (x₁, y₁) and (x₂, y₂): ((x₁ + x₂)/2, (y₁ + y₂)/2). Distance: √[(x₂ − x₁)² + (y₂ − y₁)²].
  • Parallel lines have equal gradients. Perpendicular lines: m₁m₂ = −1 LEARN IT.
  • Intersection of two lines: solve the equations simultaneously.
Worked example

Find the equation of the perpendicular bisector of A(−1, 3) and B(5, 7), in the form ax + by = c.

  1. midpoint M = (2, 5) B1
  2. gradient AB = (7 − 3)/(5 − (−1)) = 4/6 = ⅔, so the perpendicular gradient is −3/2 M1
  3. y − 5 = −3/2(x − 2) M1
  4. 2y − 10 = −3x + 6 ⇒ 3x + 2y = 16 A1
  5. check: M(2, 5) gives 6 + 10 = 16 ✓
Common mistakeUsing the curve's gradient for the perpendicular. In 2024 some candidates took the perpendicular gradient as ⅓ from a gradient of 3 (the reciprocal, without the minus sign). The perpendicular gradient is −1/m.
Exam tipThe perpendicular bisector of AB is the set of points equidistant from A and B. It passes through the midpoint with gradient −1/m_AB. If A and B lie on a circle, it passes through the centre.

2Circles

Two forms of the equation

(x − a)² + (y − b)² = r² LEARN IT
centre (a, b), radius r (it is Pythagoras: the distance from (x, y) to the centre is r)
x² + y² + 2gx + 2fy + c = 0
centre (−g, −f), radius √(g² + f² − c)
  • From expanded to centre–radius form: complete the square in x and in y separately. This is safer than memorising the second form.
Worked example

Find the centre and radius of the circle x² + y² − 6x + 4y − 12 = 0.

  1. (x − 3)² − 9 + (y + 2)² − 4 − 12 = 0 M1
  2. (x − 3)² + (y + 2)² = 25
  3. centre (3, −2), radius 5 A1 A1
Worked example

A and B have coordinates (1, 2) and (7, 10). Find the equation of the circle with diameter AB.

  1. centre = midpoint of AB = (4, 6) B1
  2. radius = ½AB = ½√(6² + 8²) = ½ × 10 = 5 M1
  3. (x − 4)² + (y − 6)² = 25 A1
Common mistakeSign errors with the centre. (x + 2)² means the centre has x-coordinate −2, not 2.

Circle properties

PropertyHow it is used
A tangent is perpendicular to the radius at the point of contacttangent gradient = −1/(gradient of radius)
The angle in a semicircle is 90°if AB is a diameter and P is on the circle, AP ⊥ BP
The perpendicular from the centre to a chord bisects the chordthe centre lies on the perpendicular bisector of every chord
Symmetry about any diameterreflect known points in a line through the centre
xy-2246810-224680C(3, 2)P(6, 6)radiustangentgradient CP = 4/3tangent gradient = −3/4perpendicular: m₁m₂ = −1(x − 3)² + (y − 2)² = 25. The tangent at P is perpendicular to the radius CP.
At P(6, 6), the radius has gradient 4/3, so the tangent has gradient −3/4.
Worked example

The point P(6, 2) lies on the circle (x − 3)² + (y + 2)² = 25. Find the equation of the tangent at P.

  1. gradient of radius CP = (2 − (−2))/(6 − 3) = 4/3 M1
  2. tangent gradient = −3/4 M1
  3. y − 2 = −¾(x − 6) M1
  4. 4y − 8 = −3x + 18 ⇒ 3x + 4y = 26 A1
  • Circle through three points A, B, C: the centre is where the perpendicular bisectors of AB and BC meet. Then the radius is the distance from the centre to any of the points.

3Where a line meets a circle

Substitute and use the discriminant

  • Substitute the line into the circle to get a quadratic in x (or y). Solve it for the intersection points.
  • The discriminant tells you the geometry: > 0 the line is a chord (two points); = 0 it is a tangent; < 0 it misses.
Worked example

Find where the line y = 2x − 3 meets the circle (x − 3)² + (y + 2)² = 25, and the length of the chord.

  1. (x − 3)² + (2x − 3 + 2)² = 25 ⇒ (x − 3)² + (2x − 1)² = 25 M1
  2. 5x² − 10x + 10 = 25 ⇒ x² − 2x − 3 = 0 ⇒ (x − 3)(x + 1) = 0 A1
  3. points (3, 3) and (−1, −5) A1
  4. chord length = √(4² + 8²) = √80 = 4√5 B1
Worked example

Find the values of c for which the line y = 2x + c is a tangent to the circle x² + y² = 5.

  1. x² + (2x + c)² = 5 ⇒ 5x² + 4cx + (c² − 5) = 0 M1
  2. tangent ⇒ discriminant = 0: 16c² − 20(c² − 5) = 0 M1
  3. 100 − 4c² = 0 ⇒ c = ±5 A1
  4. check: the distance from the centre (0, 0) to 2x − y + c = 0 is ǀcǀ/√5 = √5 = r ✓
  • Relating graphs and equations: "Find the set of values of k for which y = x + k meets the curve" means "find k so that the discriminant is ≥ 0". See Quadratics.
Common mistakeSubstituting into the circle and forgetting to square the whole bracket: (2x − 1)² = 4x² − 4x + 1, not 4x² + 1.
Linking questionThe discriminant method is from Quadratics; perpendicular gradients give normals to curves in Differentiation; and the same circle equations in complex form are loci in Complex numbers.

Test yourself — cover the answers first

Quick check

  1. What is the gradient of a line perpendicular to 2x + 5y = 7?
  2. Find the centre and radius of (x + 1)² + (y − 4)² = 49.
  3. Write down the condition for AP ⊥ BP when AB is a diameter of a circle through P.
  4. How do you find the centre of a circle through three given points?

Exam-style

  1. The points A(2, 1) and B(8, 5) lie on a circle whose centre lies on the line y = x − 2. Find the equation of the circle.
  2. The line x + y = k is a tangent to the circle x² + y² − 4x − 6y + 9 = 0. Find the possible values of k.
  3. Find the equation of the tangent to x² + y² − 2x + 4y − 20 = 0 at the point (4, 2).
Show answers
  1. 5/2
  2. (−1, 4), 7
  3. angle in a semicircle is 90°
  4. intersect the perpendicular bisectors of two chords
  5. perpendicular bisector of AB: 3x + 2y = 21; with y = x − 2: centre (5, 3); r² = 13; (x − 5)² + (y − 3)² = 13
  6. centre (2, 3), r = 2; ǀ5 − kǀ/√2 = 2 ⇒ k = 5 ± 2√2
  7. centre (1, −2); radius gradient 4/3; 3x + 4y = 20
Practise this chapter
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Independent notes written by Riyaz Sir. Not produced, endorsed or approved by Cambridge University Press & Assessment. Syllabus codes 9709 and 9702 refer to the Cambridge International AS & A Level syllabuses.