Topic 3 covers straight lines (gradients, parallel and perpendicular, the equation y − y₁ = m(x − x₁)), circles (completing the square to find centre and radius, tangents, chords and the angle in a semicircle), and parametric equations — including converting to Cartesian form and the parametric circle.
Work through each screen, answer the questions as you go (some are wordy, most are calculations) and collect ⭐ stars. Watch for the Exam trap notes — that is where the marks get thrown away. Press Start when you are ready.
Everything flows from the gradient.
Gradient AB = (10 − 2)/(5 − 1) = 8/4 = 2. Midpoint = ((1+5)/2, (2+10)/2) = (3, 6).
Perpendicular gradient = −1/2. So y − 6 = −½(x − 3) ⇒ y = −½x + 7.5.
Check: the point (3, 6) satisfies it, and −½ × 2 = −1 ✓
Exam trap: a vertical line has no gradient (undefined) and equation x = k — the perpendicular to it is horizontal, y = c. The m₁m₂ = −1 rule cannot be used there.
Midpoint (3, 6); perpendicular gradient −½.
y − 6 = −½(x − 3). At x = 0: y = 6 + 1.5 = 7.5.
Δx = 5 − (−1) = 6, Δy = −4 − 4 = −8.
AB = √(6² + (−8)²) = √(36 + 64) = √100 = 10.
A circle is the set of points a fixed distance r from a centre (a, b) — that is Pythagoras.
Group and complete the square: (x² − 6x) + (y² + 4y) = 12
(x − 3)² − 9 + (y + 2)² − 4 = 12
(x − 3)² + (y + 2)² = 12 + 9 + 4 = 25
Centre (3, −2), radius 5. Check the point (7, 1): (7−3)² + (1+2)² = 16 + 9 = 25 ✓ so it lies on the circle.
Exam trap: the radius is √25 = 5, not 25. And the centre of (x − 3)² + (y + 2)² is (3, −2) — flip the signs inside the brackets.
(x − 3)² − 9 + (y + 2)² − 4 − 12 = 0
(x − 3)² + (y + 2)² = 25, so r = √25 = 5, centre (3, −2).
Three circle facts are examined again and again:
Gradient CP = (1 − (−2))/(7 − 3) = 3/4.
Tangent gradient = −1 ÷ (3/4) = −4/3.
Tangent: y − 1 = −4/3 (x − 7) ⇒ 3y + 4x = 31.
Gradient CP = (1 + 2)/(7 − 3) = 3/4.
Tangent ⊥ radius ⇒ gradient = −4/3 ≈ −1.33.
The circle is x² + y² = 25 (centre O, radius 5). Tap a point, then say where it lies. Compare x² + y² with 25.
Instead of one equation linking x and y, give both in terms of a parameter t (or θ). Each value of t gives one point on the curve.
From the first equation t = x/2.
Substitute: y = (x/2)² − 3 = x²/4 − 3 — a parabola.
When x = 10: t = 5, so y = 25 − 3 = 22. (Cartesian check: 100/4 − 3 = 25 − 3 = 22 ✓)
The parametric circle. x = a + r cos θ, y = b + r sin θ gives the circle centre (a, b), radius r, because (x − a)² + (y − b)² = r²(cos²θ + sin²θ) = r². The identity cos²θ + sin²θ ≡ 1 is the standard tool for eliminating a trigonometric parameter.
Exam trap: the parameter often restricts the domain. If x = t² then x ≥ 0, so only part of the Cartesian curve is actually traced. Always state the domain when you convert.
x = 2t = 10 ⇒ t = 5.
y = t² − 3 = 25 − 3 = 22.
Cartesian form y = x²/4 − 3 gives the same: 100/4 − 3 = 22 ✓
Substitute the line into the circle, form a quadratic, then read the discriminant — this links Topics 2 and 3.
x² + (x + 1)² = 25 ⇒ 2x² + 2x + 1 = 25 ⇒ 2x² + 2x − 24 = 0 ⇒ x² + x − 12 = 0.
Δ = 1² − 4(1)(−12) = 49 > 0, so two intersections: (x + 4)(x − 3) = 0 ⇒ x = −4 or x = 3.
Points: (−4, −3) and (3, 4). Check (3, 4): 9 + 16 = 25 ✓
Alternative for tangency: the perpendicular distance from the centre to the line equals the radius. Both methods are accepted, but the discriminant method also hands you the coordinates.
Tap a card on the left, then its match on the right.
A typical 9MA0 circle question chains everything:
The circle C has equation x² + y² − 6x + 4y − 12 = 0. The line l is the tangent at P(7, 1). Find where l crosses the x-axis.
1. Complete the square: centre (3, −2), r = 5.
2. Check P is on C: 16 + 9 = 25 ✓
3. Gradient CP = 3/4 ⇒ tangent gradient −4/3.
4. l: y − 1 = −4/3(x − 7) ⇒ 3y − 3 = −4x + 28 ⇒ 4x + 3y = 31.
5. On the x-axis y = 0 ⇒ x = 31/4 = 7.75. So l meets the x-axis at (7.75, 0).
Marks are lost by: forgetting to take the square root for r; using the gradient of the radius as the tangent gradient; and rounding too early. Keep fractions exact until the last line.
Lines: m = Δy/Δx · y − y₁ = m(x − x₁) · parallel m₁ = m₂ · perpendicular m₁m₂ = −1
Circle: (x − a)² + (y − b)² = r²; complete the square from the expanded form
Tangent: perpendicular to the radius at the point of contact
Chord / semicircle: perpendicular bisector of a chord passes through the centre; angle in a semicircle is 90°
Parametric: eliminate t to get the Cartesian equation; x = a + r cos θ, y = b + r sin θ is a circle
Topic 3 is where algebra and geometry meet — and it returns in parametric differentiation in Topic 7. Press Finish to see your score.
You have worked through Coordinate geometry for Edexcel A-level Mathematics (9MA0). 🎉
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