Mini-Lesson
Coordinate geometry
This mini-lesson covers straight lines (gradient, midpoint, distance, parallel and perpendicular), the equation of a circle in both forms, tangents and normals , the three circle properties examiners love, and an introduction to parametric equations .
Where this sits: Unit AS 1: Pure Mathematics — coordinate geometry in the (x, y) plane, including the straight line and the circle. Unit A2 1 extends this to parametric equations of curves.
Work through each screen, answer the questions as you go (some are reasoning, some are calculations) and collect ⭐ stars. Press Start when you are ready.
Straight lines
Gradient, midpoint and distance
m = (y₂ − y₁)/(x₂ − x₁) · midpoint = ((x₁+x₂)/2, (y₁+y₂)/2)distance AB = √[(x₂ − x₁)² + (y₂ − y₁)²] — Pythagoras in disguise
Line forms you must know: y − y₁ = m(x − x₁) (point–gradient), y = mx + c , and ax + by + c = 0 (needed when a gradient is fractional).
Worked example — distance
A(2, −1), B(7, 11). Δx = 5, Δy = 12.
AB = √(5² + 12²) = √(25 + 144) = √169 = 13
Calculate
Distance between two points
1 Find the exact distance between A(2, −1) and B(7, 11).
units
Check ✓
Hint: Δx = 7 − 2 = 5 and Δy = 11 − (−1) = 12. Distance = √(5² + 12²) = √169.
Straight lines
Parallel and perpendicular
parallel: m₁ = m₂ · perpendicular: m₁ × m₂ = −1so the perpendicular gradient is the negative reciprocal: m₂ = −1/m₁
Worked example — perpendicular bisector
A(−1, 4), B(5, −2). Gradient AB = (−2 − 4)/(5 − (−1)) = −6/6 = −1.
Perpendicular gradient = −1/(−1) = 1 . Midpoint = ((−1 + 5)/2, (4 + (−2))/2) = (2, 1) .
y − 1 = 1(x − 2) ⇒ y = x − 1 , so the y-intercept is −1.
Special cases: a horizontal line (m = 0) is perpendicular to a vertical line (gradient undefined) — the product rule m₁m₂ = −1 breaks down there, so argue geometrically.
Calculate
Perpendicular bisector
2 Find the perpendicular bisector of the line joining A(−1, 4) and B(5, −2). State its y-intercept .
Check ✓
Hint: Gradient AB = −6/6 = −1, so the bisector has gradient 1. Midpoint is (2, 1). Then y − 1 = 1(x − 2) gives y = x − 1.
Circles
The equation of a circle
(x − a)² + (y − b)² = r²centre (a, b), radius r · expanded form: x² + y² + 2gx + 2fy + c = 0
To go from the expanded form to the centre and radius, complete the square in x and in y.
Worked example — x² + y² − 6x + 4y − 12 = 0
(x² − 6x) + (y² + 4y) = 12
(x − 3)² − 9 + (y + 2)² − 4 = 12
(x − 3)² + (y + 2)² = 25 ⇒ centre (3, −2) , radius 5
If the right-hand side comes out negative , there is no circle . Always check r² > 0.
Calculate
Centre and radius
3 Find the radius of the circle x² + y² − 6x + 4y − 12 = 0.
units
Check ✓
Hint: Complete the square: (x − 3)² + (y + 2)² = 12 + 9 + 4 = 25, so r = √25.
Circles
Tangents, normals and the three key properties
A tangent meets the circle once and is perpendicular to the radius at the point of contact.
The perpendicular from the centre to a chord bisects the chord .
The angle in a semicircle is a right angle — so if ∠APB = 90°, AB is a diameter .
Worked example — tangent gradient
Circle centre C(3, −2), radius 5. Point P(6, 2): check (6 − 3)² + (2 + 2)² = 9 + 16 = 25 ✓ so P is on the circle.
Gradient of radius CP = (2 − (−2))/(6 − 3) = 4/3.
Tangent ⟂ radius ⇒ gradient of tangent = −1 ÷ (4/3) = −3/4 = −0.75
Calculate
Gradient of a tangent
4 The circle has centre C(3, −2) and radius 5. The point P(6, 2) lies on it. Find the gradient of the tangent at P.
Check ✓
Hint: Gradient of the radius CP = (2 + 2)/(6 − 3) = 4/3. The tangent is perpendicular, so its gradient is the negative reciprocal, −3/4.
Quick check
Is it a diameter?
? P lies on a circle with A and B also on the circle, and ∠APB = 90°. What follows?
AB is a tangent ❌
AB is a diameter ✅
P is the centre ❌
AP = PB always ❌
Sort it
Inside, on, or outside?
The circle is (x − 2)² + (y + 1)² = 25 (centre (2, −1), r = 5). Tap a point, then tap where it lies.
A2 skill
Parametric equations
Instead of y = f(x), give x and y each in terms of a parameter t. To convert to Cartesian form, eliminate t .
Worked example — x = 2t, y = t² − 1
From the first equation t = x/2.
Substitute: y = (x/2)² − 1 = x²/4 − 1 — a parabola.
At x = 6: t = 3, so y = 3² − 1 = 8 (and the Cartesian form agrees: 36/4 − 1 = 8 ✓).
Circle in parametric form: x = a + r cos t, y = b + r sin t. Squaring and adding rebuilds (x − a)² + (y − b)² = r², since cos²t + sin²t = 1.
Calculate
Parametric to Cartesian
5 A curve has parametric equations x = 2t, y = t² − 1. Find y when x = 6 .
Check ✓
Hint: x = 2t = 6 gives t = 3, so y = 3² − 1 = 8. (Or eliminate t: y = x²/4 − 1 = 36/4 − 1 = 8.)
Quick check
Perpendicular test
? Which line is perpendicular to 2x + 3y = 6?
y = −(2/3)x + 1 ❌
y = (3/2)x − 4 ✅
y = (2/3)x + 5 ❌
y = −(3/2)x ❌
Match it
Match the circle facts
Tap the item on the left, then its value on the right.
Quick check
Chord bisector
? A chord of a circle has midpoint M. Which statement is always true?
The line from the centre to M is perpendicular to the chord ✅
M is the centre of the circle ❌
The chord is a diameter ❌
The tangent at M passes through the centre ❌
Quick check
Intersection of line and circle
? Substituting a line into a circle gives a quadratic with discriminant b² − 4ac < 0. What does that mean geometrically?
The line is a tangent ❌
The line cuts the circle twice ❌
The line misses the circle entirely ✅
The line passes through the centre ❌
Examiner traps
Circle and line pitfalls
Sign of the centre. (x − 3)² + (y + 2)² = 25 has centre (3, −2 ) — the signs flip.Radius vs r². The right-hand side is r², so r = √25 = 5, not 25.Tangent gradient. Take the negative reciprocal of the radius gradient — do not reuse the radius gradient.Perpendicular bisector needs BOTH the midpoint and the perpendicular gradient. Many candidates supply only one.Discriminant of the substituted quadratic tells you 2 roots (cuts), 0 (tangent), none (misses) — a fast, mark-scoring method.
Recap
The big ideas to know
Lines: m = Δy/Δx · midpoint · distance = √(Δx² + Δy²) · m₁m₂ = −1 for perpendicular
Circle: (x − a)² + (y − b)² = r²; complete the square from the expanded form
Tangent ⟂ radius ; the perpendicular from the centre bisects a chord; angle in a semicircle = 90°
Line meets circle: substitute and read the discriminant — 2 roots (cuts), 1 (tangent), none (misses)
Parametric: eliminate the parameter to recover the Cartesian equation
Circle work is a favourite CCEA source of multi-step problems combining gradients, midpoints and the discriminant. Press Finish to see your score.
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