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OCR A-level Mathematics A (H240) ยท 1.03 Coordinate Geometry in the xโ€“y Plane
Mini-Lesson

Coordinate geometry

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.

Straight lines Circles Parametric curves every circle question is really a triangle question in disguise

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.

Coordinate geometry · lines

Straight lines

y − y₁ = m(x − x₁)  ·  m = (y₂ − y₁)/(x₂ − x₁)parallel lines: mโ‚ = mโ‚‚  ยท  perpendicular lines: mโ‚ ร— mโ‚‚ = โˆ’1, i.e. mโ‚‚ = โˆ’1/mโ‚
Worked example โ€” A(1, 2) and B(5, 10)

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.

Calculate

Your turn โ€” perpendicular gradient

1A(1, 2) and B(5, 10). Find the gradient of a line perpendicular to AB.
m =
Hint: gradient of AB = (10 − 2)/(5 − 1) = 2. Perpendicular gradient = −1/2.
Coordinate geometry · circles

The equation of a circle

(x − a)² + (y − b)² = r²centre (a, b), radius r โ€” this is just Pythagoras on the radius

Exam circles usually arrive expanded. Get them back into completed-square form.

Worked example โ€” x² + y² + 6x − 4y − 3 = 0

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.

Calculate

Your turn โ€” radius of a circle

2Find the radius of the circle x² + y² + 6x − 4y − 3 = 0.
r =
Hint: complete the square → (x + 3)² + (y − 2)² = 9 + 4 + 3 = 16, so r = √16.
Quick check

Read off the circle

?A circle has equation (x − 2)² + (y + 5)² = 9. What is its centre and radius?
Coordinate geometry · tangents

Tangents and the radius

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).

Worked example โ€” tangent length from an external point

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.

Calculate

Your turn โ€” tangent length

3A circle has centre C(3, −2) and radius 5. Find the length of the tangent from the point P(10, 2), to 3 significant figures.
units
Hint: CP² = 7² + 4² = 65. Tangent length = √(65 − 5²) = √40.
Calculate

Your turn โ€” diameter to radius

4A(1, 3) and B(7, 11) are the ends of a diameter of a circle. Find the circle’s radius.
r =
Hint: AB = √((7 − 1)² + (11 − 3)²) = √(36 + 64) = √100 = 10. The radius is half the diameter.
Quick check

Tangent and radius

?P lies on a circle with centre C. What is always true about the tangent to the circle at P?
Coordinate geometry · parametric

Parametric equations

Instead of linking y to x directly, both are given in terms of a parameter t. To get the Cartesian equation, eliminate the parameter.

Worked example โ€” x = t + 1, y = t² − 2t

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. ✓

Worked example โ€” a circle in parametric form

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.

Calculate

Your turn โ€” parametric to Cartesian

5A curve has parametric equations x = t + 1, y = t² − 2t. Find the value of y when x = 5.
y =
Hint: x = 5 ⇒ t = 4. Then y = 4² − 2(4) = 16 − 8. (Or use the Cartesian form y = x² − 4x + 3.)
Quick check

Eliminating the parameter

?A curve is given by x = 2cosθ, y = 2sinθ. Which identity converts this to Cartesian form?
Sort it

Parallel, perpendicular or neither?

Each card is a straight line. Sort each one by its relationship to the line y = 2x โˆ’ 1.

โˆฅ Parallel

โŠฅ Perpendicular

โœ–๏ธ Neither

Match it

Centre and radius

Tap a circle equation on the left, then its centre and radius on the right.

Circle
Centre and radius
Quick check

Line through a point

?Which is the equation of the line through (2, 5) with gradient 3?
Coordinate geometry · problem solving

Chords, bisectors and intersections

Two more circle techniques OCR examines every year.

1 โ€” the perpendicular bisector of a chord passes through the centre

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.

2 โ€” where does a line meet a circle?

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.

Recap

The big ideas to know

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.

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