Lesson 3: Coordinate Geometry

Coordinate geometry turns algebra into pictures and pictures into algebra — the bridge Descartes built between equations and shapes. At A-level it goes beyond GCSE straight lines into circles, tangents, and the parametric thinking that powers calculus and mechanics. This lesson covers the A-level toolkit: gradient, distance, and midpoint in fluent use; the circle equation (x − a)² + (y − b)² = r²; tangents and normals; and intersection problems solved by simultaneous equations. Every result here is proved, not just stated — this is geometry with reasons.

Lines — gradient, distance, midpoint

The gradient between (x₁, y₁) and (x₂, y₂) is (y₂ − y₁)/(x₂ − x₁): rise over run. Parallel lines share gradients; perpendicular lines have gradients multiplying to −1 (m₁m₂ = −1) — proved by rotating a gradient triangle through 90°. The distance formula d = √((x₂ − x₁)² + (y₂ − y₁)²) is Pythagoras in disguise; the midpoint is ((x₁ + x₂)/2, (y₁ + y₂)/2). These three formulas solve a surprising range of problems.

Worked example: A(2, 5), B(8, −3). The midpoint M is (5, 1). Gradient AB = (−3 − 5)/(8 − 2) = −8/6 = −4/3. The perpendicular bisector passes through M with gradient 3/4 (since (−4/3)(3/4) = −1): equation y − 1 = (3/4)(x − 5), i.e. 3x − 4y − 11 = 0. This construction — perpendicular bisector — is the key to circle centres in the next section.

Exam-style: find the equation of the line through (4, 7) perpendicular to 2x + 5y = 10. Rearrange: y = −2x/5 + 2, so gradient −2/5; perpendicular gradient 5/2. Line: y − 7 = (5/2)(x − 4) → 5x − 2y − 6 = 0. Always convert to a comparable form before reading gradients — the common error is reading −2/5 as 2/5.

The circle — equation and features

A circle with centre (a, b) and radius r is the set of points at distance r from (a, b): by the distance formula, (x − a)² + (y − b)² = r². This is the definition turned into an equation — every A-level circle problem starts here. Example: centre (3, −2), radius 5: (x − 3)² + (y + 2)² = 25. A point is on the circle iff its coordinates satisfy the equation: does (6, 2) lie on it? (6 − 3)² + (2 + 2)² = 9 + 16 = 25 ✓ — yes.

Worked example (circle from a diameter). Find the equation of the circle with diameter endpoints P(1, 2) and Q(7, 10). The centre is the midpoint: ((1 + 7)/2, (2 + 10)/2) = (4, 6). The radius is half of PQ = √(36 + 64) = 10, so r = 5. Equation: (x − 4)² + (y − 6)² = 25. Diameter problems are midpoint + distance in disguise.

Completing the square converts expanded form to centre–radius form. Given x² + y² − 6x + 8y − 11 = 0: group x's and y's: (x² − 6x) + (y² + 8y) = 11 → (x − 3)² − 9 + (y + 4)² − 16 = 11 → (x − 3)² + (y + 4)² = 36. Centre (3, −4), radius 6. If the right-hand side is negative, no real points satisfy the equation (an imaginary circle); if zero, a single point. Examiners test this: 'show that x² + y² + 2x − 4y + 8 = 0 represents no real circle' — completing the square gives (x + 1)² + (y − 2)² = −3, impossible.

The general second-degree equation represents a circle only if the x² and y² coefficients are equal and there is no xy term. So 2x² + 2y² − 8x + 6y = 0 is a circle (divide by 2 first), but 3x² + 2y² = 12 is an ellipse, not on the A-level syllabus as such — recognising the difference is a quick exam mark.

Tangents and normals to circles

The tangent at a point on a circle is perpendicular to the radius to that point — the key theorem. For the circle x² + y² = 25 at point (3, 4): the radius gradient is 4/3, so the tangent gradient is −3/4, and the tangent equation is y − 4 = (−3/4)(x − 3), i.e. 3x + 4y = 25. Notice the elegant pattern: for x² + y² = r² at (x₁, y₁), the tangent is xx₁ + yy₁ = r². This 'T = 0' form generalises to all conics and is worth memorising.

Worked example: find the tangents from (10, 2) to the circle x² + y² = 25. Let the tangent touch at (x₁, y₁): then x₁² + y₁² = 25 and the tangent xx₁ + yy₁ = 25 passes through (10, 2): 10x₁ + 2y₁ = 25. Solve simultaneously: from the linear equation y₁ = (25 − 10x₁)/2; substitute into the circle to get a quadratic in x₁ with two solutions — the two points of contact. The algebra is heavy but routine; the geometry tells you what to expect (two tangents from an exterior point, one from a point on the circle, none from inside).

The normal at a point passes through the centre (it is the radius line extended). For the circle above at (3, 4), the normal is y = (4/3)x. Tangent–normal pairs appear in optimisation and reflection problems: the shortest path from a point to a circle follows the normal through the centre.

Intersections — lines with lines, lines with circles

Intersections are simultaneous equations. Line with line: solve 2x + 3y = 7 and x − y = 1 — substitute or eliminate to get (2, 1). Line with circle: substitute the linear equation into the circle to get a quadratic; its discriminant tells the story: two solutions (secant — line cuts the circle), one (tangent — touches), none (misses). This unifies algebra and geometry beautifully.

Worked example: does the line y = 2x + 1 intersect the circle x² + y² = 10? Substitute: x² + (2x + 1)² = 10 → 5x² + 4x − 9 = 0. Discriminant: 16 + 180 = 196 > 0 — two intersections (secant). Solving: x = (−4 ± 14)/10, so x = 1 or x = −1.8; points (1, 3) and (−1.8, −2.6). For tangency, set the discriminant to zero and solve for the unknown parameter — a classic 'find k such that the line is a tangent' problem.

Example: find k such that y = x + k is tangent to x² + y² = 8. Substitute: x² + (x + k)² = 8 → 2x² + 2kx + (k² − 8) = 0. Tangent ⟺ discriminant zero: 4k² − 8(k² − 8) = 0 → −4k² + 64 = 0 → k² = 16 → k = ±4. Two parallel tangents — sketch to see why. The discriminant-as-geometric-detector is one of A-level's most satisfying ideas.

Proofs and loci

Coordinate geometry proves theorems: the perpendicular-bisector construction shows the circumcentre of a triangle is equidistant from all vertices (it lies on all three bisectors — prove two bisectors meet at a point equidistant from the vertices, hence on the third). The midpoint theorem, the diagonals of a rectangle, the circle theorems — all yield to coordinates. A-level occasionally asks for such proofs: set up coordinates cleverly (e.g. place a rectangle at (0, 0), (a, 0), (a, b), (0, b)) and let algebra do the work.

A locus is a set of points satisfying a condition — an equation waiting to be found. Example: find the locus of points P such that PA = 2·PB, with A(0, 0), B(3, 0). Let P = (x, y): √(x² + y²) = 2√((x − 3)² + y²) → x² + y² = 4((x − 3)² + y²) → 3x² − 24x + 36 + 3y² = 0 → x² − 8x + 12 + y² = 0 → (x − 4)² + y² = 4. The locus is a circle (Apollonius circle) centre (4, 0) radius 2. Locus problems are equation-derivation in pure form.

In context: a delivery company in Birmingham wants the set of points within 5 km of two depots — intersecting circles; a mobile mast's coverage is a disc (x − a)² + (y − b)² ≤ r². Coordinate geometry is how the abstract becomes applicable — every GPS fix is an intersection of loci.

Common mistakes

• Reading the gradient straight from ax + by = c. For 2x + 5y = 10 the gradient is −2/5, not 2/5 and not −5/2. Rearrange to y = mx + c first, every time.
• Sign errors completing the square. x² + 8y... the centre's y-coordinate is −4 when the bracket is (y + 4)². Write the centre as (a, b) from (x − a)² + (y − b)² and take the opposite signs.
• Using the radius gradient as the tangent gradient. The tangent is perpendicular to the radius: m_tangent = −1/m_radius. Forgetting the negative reciprocal is the classic circle error.
• Forgetting both tangents. 'Find k' tangency problems usually have two answers (k = ±4 above). If your quadratic gives one value, check whether the geometry expects a symmetric pair.
• Stopping at the discriminant. A positive discriminant says the line cuts the circle; if the question asks for the points, you must still solve the quadratic and find both coordinates of each point.

Exam Experience

Questions in the style of the A-level pure paper. Attempt them before reading the mark scheme.

1. A circle C has equation x² + y² − 4x + 6y = 12. (a) Find the centre and radius of C. [3] (b) Show that the point P(5, 1) lies on C. [1] (c) Find the equation of the tangent to C at P, giving your answer in the form ax + by = c. [3]

Mark scheme.

• 1(a) Complete the square: (x − 2)² − 4 + (y + 3)² − 9 = 12 (M1) for both completed squares. → (x − 2)² + (y + 3)² = 25, so centre (2, −3) (A1), radius 5 (A1).
• 1(b) Substitute: 5² + 1² − 4(5) + 6(1) = 25 + 1 − 20 + 6 = 12 (M1) — equals the right-hand side, so P lies on C (A1). In a 'show that' part, the substitution must be shown in full, not asserted.
• 1(c) Gradient of radius CP = (1 − (−3))/(5 − 2) = 4/3 (M1). Tangent gradient = −3/4 (negative reciprocal) (M1). Tangent: y − 1 = −(3/4)(x − 5) → 4y − 4 = −3x + 15 → 3x + 4y = 19 (A1).