Most tested Papers 1 & 2

Coordinate Geometry: Straight Lines, Circles and Circle Theorems (TMUA)

Points, lines and circles in the (x, y)-plane are handled through gradients, distances and equations, then sharpened by the classical circle theorems. TMUA rewards picking the standard circle form, the perpendicular-from-centre trick or one theorem to sidestep heavy algebra and reach an exact answer by hand.

Part of the TMUA syllabus — revision for the Test of Mathematics for University Admission (TMUA), the UAT-UK maths test used by Cambridge, Oxford, Imperial, UCL, LSE, Warwick and Durham. No calculator; multiple choice.

Key points

  • The line through (x1, y1) with gradient m is y - y1 = m(x - x1); from two points the gradient is m = (y2 - y1)/(x2 - x1).
  • Parallel lines have equal gradients; perpendicular lines have gradients whose product is -1, so one is the negative reciprocal of the other.
    m1 = m2
    m1 × m2 = -1
  • Distance between two points is d = √((x2 - x1)2 + (y2 - y1)2), and the midpoint is ((x1 + x2)/2, (y1 + y2)/2).
  • A circle written as (x - a)2 + (y - b)2 = r2 has centre (a, b) and radius r read off directly.
  • For the general form x2 + y2 + cx + dy + e = 0, completing the square gives centre (-c/2, -d/2) and r2 = (c/2)2 + (d/2)2 - e; a genuine circle needs r2 > 0.
  • A tangent meets the radius at its point of contact at a right angle, so mtangent = -1/mradius; this is the fastest route to a tangent's equation.
  • The perpendicular from the centre bisects a chord, so (half-chord)2 + (distance from centre)2 = r2 gives chord lengths without trigonometry.
  • Core circle theorems: angle in a semicircle is 90 degrees, angles in the same segment are equal, opposite angles of a cyclic quadrilateral sum to 180 degrees, and the tangent-chord angle equals the angle in the alternate segment.

Diagram

centrertangenttangent meets radius at 90 degrees
A tangent to a circle meets the radius at the point of contact at a right angle.

Formulae

y - y1 = m(x - x1)

Equation of a straight line through a known point with gradient m.

m1 × m2 = -1

Test or impose that two non-vertical lines are perpendicular (parallel lines instead have m1 = m2).

d = √((x2 - x1)2 + (y2 - y1)2)

Distance between two points, or a radius measured from the centre to a point on the circle.

(x - a)2 + (y - b)2 = r2

Circle with centre (a, b) and radius r; read the centre and radius straight off.

centre = (-c/2, -d/2), r2 = (c/2)2 + (d/2)2 - e

Convert the general form x2 + y2 + cx + dy + e = 0 into centre and radius.

half-chord = √(r2 - h2)

Length of half a chord that lies at perpendicular distance h from the centre.

Definitions

Chord
A straight segment joining two points on a circle. The perpendicular drawn from the centre to a chord always bisects it, which lets you split it into two right-angled triangles.
Tangent
A line meeting a circle at exactly one point; it is perpendicular to the radius drawn to that point of contact, so radius gradient times tangent gradient equals -1.
Cyclic quadrilateral
A four-sided figure whose four vertices all lie on one circle; its opposite angles add up to 180 degrees.
Alternate segment theorem
The angle between a tangent and a chord at the point of contact equals the inscribed angle that the same chord subtends in the alternate (opposite) segment.

Worked examples

1

A circle has equation x2 + y2 - 6x + 4y - 12 = 0. Find its centre and radius, and find the equation of the tangent to the circle at the point (7, 1).

  1. 1

    Complete the square in x and y:

    (x-3)2 + (y+2)2 = 25.

  2. 2

    Read off from the standard form:

    centre = (3, -2), r = 5
  3. 3

    Confirm (7, 1) is on the circle:

    (7-3)2 + (1+2)2 = 25.

  4. 4

    Gradient of the radius to (7, 1):

    mr = (1+2)/(7-3) = 3/4
  5. 5

    Tangent is perpendicular to the radius:

    mt = -4/3
  6. 6

    Line through (7, 1):

    y - 1 = -(4/3) × (x - 7)
  7. 7

    Clear fractions and tidy:

    4x + 3y = 31

Answer: Centre (3, -2), radius 5; the tangent at (7, 1) is 4x + 3y = 31.

2

A(1, 2) and B(7, 10) are the ends of a diameter of a circle. Find the equation of the circle, then verify that C(8, 9) lies on it and that angle ACB is a right angle.

  1. 1

    Centre is the midpoint of the diameter AB:

    centre = (4, 6)
  2. 2

    Radius is the distance from the centre to A:

    r = √(32 + 42) = 5
  3. 3

    Equation of the circle:

    (x-4)2 + (y-6)2 = 25.

  4. 4

    Check C(8, 9) lies on it:

    (8-4)2 + (9-6)2 = 25.

  5. 5

    Gradient of CA:

    mCA = (2-9)/(1-8) = 1
  6. 6

    Gradient of CB:

    mCB = (10-9)/(7-8) = -1
  7. 7

    Product of gradients:

    mCA × mCB = -1, so CA is perpendicular to CB

Answer: Circle (x-4)2 + (y-6)2 = 25; C(8, 9) lies on it and angle ACB = 90 degrees, matching the angle-in-a-semicircle theorem since AB is a diameter.

Common mistakes

  • ×Treating the perpendicular gradient as the negative rather than the negative reciprocal: if m = 2/3 the perpendicular gradient is -3/2, not -2/3.
  • ×Sign and formula slips in the general circle form: the centre is (-c/2, -d/2) with signs flipped, and r2 = (c/2)2 + (d/2)2 - e (subtract e, do not add it).
  • ×Reading the right-hand side of (x-a)2 + (y-b)2 = r2 as the radius: if it equals 16 the radius is 4, not 16.
  • ×Using the angle-in-a-semicircle result without first checking that the chord is actually a diameter (it must pass through the centre).
  • ×Assuming any line to a chord bisects it: only the perpendicular drawn from the CENTRE bisects a chord.

No-calculator tips

  • ✓Watch for Pythagorean triples (3-4-5, 5-12-13, 8-15-17) in distance and radius work so the square root collapses to a whole number.
  • ✓Check perpendicularity by inspection: multiply the two gradients and look for -1; a vertical line (undefined gradient) and a horizontal line (gradient 0) are perpendicular automatically.
  • ✓Before grinding algebra, ask whether a circle theorem shortcuts it: a right angle hints at a diameter or a tangent-radius pair, and equal angles hint at the same-segment rule.
  • ✓To classify a point as inside, on, or outside a circle, compare (x-a)2 + (y-b)2 against r2 rather than computing distances explicitly.
  • ✓Complete the square mentally: halve the coefficient of x, and the x-coordinate of the centre is minus that half; do the same for y to convert general form fast.

Test yourself

Original practice questions, no calculator. Work each out before revealing the answer.

Q1.The points A(2, -1) and B(6, 7) are given. What is the equation of the perpendicular bisector of AB?

  • A. y = 2x - 5
  • B. y = x/2 + 1
  • C. y = -x/2 + 5
  • D. y = -x/2
Show answer

Answer: C — y = -x/2 + 5

The midpoint of AB is (4, 3) and the gradient of AB is 8/4 = 2, so the perpendicular bisector has gradient -1/2, giving y - 3 = -1/2(x - 4), i.e. y = -x/2 + 5. Option 'y = 2x - 5' forgets to take the negative reciprocal; 'y = -x/2' uses point A instead of the midpoint.

Q2.A circle has equation x2 + y2 - 6x + 8y - 11 = 0. What are its centre and radius?

  • A. Centre (3, -4), radius 6
  • B. Centre (-3, 4), radius 6
  • C. Centre (3, -4), radius √(14)
  • D. Centre (3, -4), radius 36
Show answer

Answer: A — Centre (3, -4), radius 6

Completing the square gives (x - 3)2 + (y + 4)2 = 9 + 16 + 11 = 36, so the centre is (3, -4) and the radius is √(36) = 6. The √(14) trap comes from writing 9 + 16 - 11 (mishandling the -11), and radius 36 forgets to square-root.

Q3.The point P(5, 5) lies on a circle with centre C(1, 2). What is the equation of the tangent to the circle at P?

  • A. 3x - 4y = -5
  • B. 4x + 3y = 35
  • C. 4x - 3y = 5
  • D. 4x + 3y = 10
Show answer

Answer: B — 4x + 3y = 35

The tangent is perpendicular to radius CP, whose gradient is (5 - 2)/(5 - 1) = 3/4, so the tangent gradient is -4/3; through P(5, 5) this gives 4x + 3y = 35. '3x - 4y = -5' wrongly uses the radius gradient, and '4x + 3y = 10' substitutes centre C rather than P.

Read this topic in the official UAT-UK TMUA content specification →

Keep going

All TMUA topics