Mini-Lesson

Coordinate geometry

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.

Coordinate geometry · lines

Straight lines

Everything flows from the gradient.

m = (y₂ − y₁)/(x₂ − x₁) · y − y₁ = m(x − x₁)parallel ⇒ m₁ = m₂ · perpendicular ⇒ m₁m₂ = −1

Distance: AB = √[(x₂ − x₁)² + (y₂ − y₁)²] — Pythagoras in disguise.
Midpoint: ((x₁ + x₂)/2, (y₁ + y₂)/2).
A perpendicular bisector passes through the midpoint with gradient −1/m.

Worked example — perpendicular bisector of A(1, 2) and B(5, 10)

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.

Calculate

Your turn — perpendicular bisector

1A is (1, 2) and B is (5, 10). The perpendicular bisector of AB crosses the y-axis at (0, c). Find c.

Set-up

Midpoint (3, 6); perpendicular gradient −½.

y − 6 = −½(x − 3). At x = 0: y = 6 + 1.5 = 7.5.

c =

Hint: Gradient AB = 2, so the bisector has gradient −½ and passes through the midpoint (3, 6).

Calculate

Your turn — distance

2Find the exact length of AB where A is (−1, 4) and B is (5, −4).

Set-up

Δx = 5 − (−1) = 6, Δy = −4 − 4 = −8.

AB = √(6² + (−8)²) = √(36 + 64) = √100 = 10.

Hint: Δx = 6, Δy = −8. Use √(Δx² + Δy²).

Coordinate geometry · circles

The equation of a circle

A circle is the set of points a fixed distance r from a centre (a, b) — that is Pythagoras.

(x − a)² + (y − b)² = r²expanded form: x² + y² + 2gx + 2fy + c = 0 — complete the square to recover the centre and radius

Worked example — x² + y² − 6x + 4y − 12 = 0

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.

Calculate

Your turn — radius of a circle

3Find the radius of the circle x² + y² − 6x + 4y − 12 = 0.

Set-up

(x − 3)² − 9 + (y + 2)² − 4 − 12 = 0

(x − 3)² + (y + 2)² = 25, so r = √25 = 5, centre (3, −2).

Hint: Complete the square in x and in y; the right-hand side becomes r².

Quick check

Centre and radius

?What are the centre and radius of (x + 1)² + (y − 3)² = 16?

Coordinate geometry · circles

Tangents, chords and the semicircle

Three circle facts are examined again and again:

A tangent is perpendicular to the radius at the point of contact.
The perpendicular bisector of a chord passes through the centre.
The angle in a semicircle is 90° — if AB is a diameter and P is on the circle, then ∠APB = 90°.

The tangent at P is perpendicular to CP, so its gradient is the negative reciprocal.

Worked example — tangent at P(7, 1) to the circle centred at C(3, −2)

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.

Calculate

Your turn — gradient of a tangent

4The circle has centre C(3, −2). Find the gradient of the tangent at the point P(7, 1), which lies on the circle. (Give a decimal or a fraction such as -4/3.)

Set-up

Gradient CP = (1 + 2)/(7 − 3) = 3/4.

Tangent ⊥ radius ⇒ gradient = −4/3 ≈ −1.33.

Hint: Find the gradient of CP, then take the negative reciprocal.

Quick check

Circle geometry

?AB is a diameter of a circle and P is a different point on the circumference. What can you say about triangle APB?

Sort it

Inside, on, or outside?

The circle is x² + y² = 25 (centre O, radius 5). Tap a point, then say where it lies. Compare x² + y² with 25.

⭕ On the circle

🎯 Inside

🚀 Outside

Coordinate geometry · parametric

Parametric equations

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.

x = f(t), y = g(t)to convert to Cartesian: make t the subject of one equation and substitute into the other

Worked example — x = 2t, y = t² − 3

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.

Calculate

Your turn — parametric to Cartesian

5A curve has parametric equations x = 2t, y = t² − 3. Find the value of y when x = 10.

Set-up

x = 2t = 10 ⇒ t = 5.

y = t² − 3 = 25 − 3 = 22.

Cartesian form y = x²/4 − 3 gives the same: 100/4 − 3 = 22 ✓

y =

Hint: x = 10 gives t = 5.

Quick check

Eliminate the parameter

?A curve is given by x = 3 cos θ, y = 3 sin θ. What is its Cartesian equation?

Coordinate geometry · intersections

Where a line meets a circle

Substitute the line into the circle, form a quadratic, then read the discriminant — this links Topics 2 and 3.

Δ > 0 → two points (a chord)Δ = 0 → one point (a tangent) · Δ < 0 → the line misses the circle

Worked example — does y = x + 1 meet x² + y² = 25?

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.

Match it

Coordinate geometry flashcards

Tap a card on the left, then its match on the right.

Given

Result

Coordinate geometry · exam technique

Putting it together

A typical 9MA0 circle question chains everything:

Full example

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.

Quick check

Straight-line facts

?Line l₁ has equation y = 4x − 1. Which line is perpendicular to l₁?

Recap

The big ideas to take away

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.

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