OCR section 1.03 covers straight lines, the circle โ including completing the square to find its centre and radius, and the fact that a tangent is perpendicular to the radius โ and, new at A-level, parametric equations.
Draw a sketch for every single question โ it is the fastest route to the method. Work through each screen, answer the questions as you go and collect ⭐ stars. Press Start when you're ready.
Gradient of AB = (10 − 2)/(5 − 1) = 8/4 = 2.
Equation of AB: y − 2 = 2(x − 1) ⇒ y = 2x.
Gradient of any line perpendicular to AB: −1/2 = −0.5.
Midpoint of AB = ((1 + 5)/2, (2 + 10)/2) = (3, 6). Length AB = √(4² + 8²) = √80 ≈ 8.94.
Careful: a horizontal line (m = 0) is perpendicular to a vertical line (gradient undefined) โ the m₁m₂ = −1 rule breaks down here, so handle that case by inspection.
Exam circles usually arrive expanded. Get them back into completed-square form.
Complete the square in x and in y:
(x + 3)² − 9 + (y − 2)² − 4 − 3 = 0
⇒ (x + 3)² + (y − 2)² = 16.
∴ centre (−3, 2), radius r = √16 = 4.
Circle facts OCR expects: the tangent is perpendicular to the radius at the point of contact; the perpendicular bisector of a chord passes through the centre; and the angle in a semicircle is 90°, so if ∠APB = 90° then AB is a diameter.
The single most useful circle fact: the tangent at P is perpendicular to the radius CP. So the gradient of the tangent is −1 ÷ (gradient of the radius).
Circle: centre C(3, −2), radius 5. External point P(10, 2). Find the length of the tangent from P.
The tangent, the radius and CP form a right-angled triangle, right-angled at the point of contact.
CP² = (10 − 3)² + (2 − (−2))² = 7² + 4² = 49 + 16 = 65.
Tangent length = √(CP² − r²) = √(65 − 25) = √40 = 6.32 (3 s.f.).
Sanity check: CP = √65 ≈ 8.06 > 5 = r, so P really is outside the circle and a tangent exists. If CP < r there is no tangent from P.
Instead of linking y to x directly, both are given in terms of a parameter t. To get the Cartesian equation, eliminate the parameter.
From the first: t = x − 1. Substitute into the second:
y = (x − 1)² − 2(x − 1) = x² − 2x + 1 − 2x + 2 = x² − 4x + 3.
Check at t = 4: the parametric form gives x = 5 and y = 16 − 8 = 8. The Cartesian form gives y = 25 − 20 + 3 = 8. ✓
x = 3 cos θ, y = 3 sin θ. Then x² + y² = 9cos²θ + 9sin²θ = 9(cos²θ + sin²θ) = 9.
So this is the circle of radius 3 centred on the origin โ eliminated using the identity sin²θ + cos²θ = 1.
Domain matters: the parameter’s range restricts the curve. x = t² (with t real) only ever gives x ≥ 0, so the Cartesian curve is only half of what it appears.
Each card is a straight line. Sort each one by its relationship to the line y = 2x โ 1.
Tap a circle equation on the left, then its centre and radius on the right.
Two more circle techniques OCR examines every year.
A chord joins A(1, 3) and B(7, 11). Midpoint = (4, 7); gradient AB = (11 − 3)/(7 − 1) = 8/6 = 4/3.
Perpendicular gradient = −3/4, so the bisector is y − 7 = −¾(x − 4). The centre lies somewhere on this line โ combine it with a second condition to pin it down.
Substitute the line into the circle and solve the resulting quadratic. The discriminant tells you everything:
Δ > 0 ⇒ the line is a secant (cuts the circle twice) · Δ = 0 ⇒ it is a tangent · Δ < 0 ⇒ it misses the circle.
Exam tip: ‘show that the line is a tangent to the circle’ almost always means substitute and show Δ = 0.
Lines: m = Δy/Δx; parallel ⇒ equal gradients; perpendicular ⇒ m₁m₂ = −1
Circle: (x − a)² + (y − b)² = r² โ complete the square to find the centre and radius
Tangent: perpendicular to the radius at the point of contact; tangent length = √(CP² − r²)
Chords: the perpendicular bisector of a chord passes through the centre
Semicircle: angle in a semicircle is 90°, so a right angle at P means AB is a diameter
Parametric: eliminate the parameter to get the Cartesian equation โ and watch the domain
That is OCR 1.03 โ and parametric curves come back in differentiation and integration. Press Finish to see your score.
You've worked through Coordinate geometry for OCR A-level Mathematics A. 🎉
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