Coordinate Geometry
Coordinate geometry on this paper mixes straight-line work with circles and simple parametric curves. You are expected to move fluently between gradient, general and parametric forms, and to justify tangency or intersection algebraically rather than from a sketch.
Straight lines
- Gradient m = (y₂ − y₁)/(x₂ − x₁); through (x₁, y₁): y − y₁ = m(x − x₁).
- ax + by + c = 0 has gradient −a/b; a perpendicular has gradient −1/m.
- Distance between points is √((Δx)² + (Δy)²); the midpoint averages the coordinates.
- Perpendicular distance from (x₁, y₁) to ax + by + c = 0 is |ax₁ + by₁ + c|/√(a² + b²).
Circles
- Centre (a, b), radius r: (x − a)² + (y − b)² = r².
- Expanded form x² + y² + 2gx + 2fy + c = 0 has centre (−g, −f) and radius √(g² + f² − c); a genuine circle needs g² + f² − c > 0.
- The perpendicular bisector of any chord passes through the centre - the quickest route to the centre when two or three points are given.
- Where a line meets a circle, substituting gives a quadratic; the discriminant decides secant (positive), tangent (zero) or no intersection (negative).
Parametric curves
For x = at², y = 2at, dy/dx = (dy/dt)/(dx/dt) = 2a/(2at) = 1/t, and the tangent at parameter t is ty = x + at². A chord joining parameters t₁ and t₂ has gradient 1/(t₁ + t₂).
Worked example
Find the tangent to x² + y² − 6x + 4y − 12 = 0 at (6, 2). The centre is (3, −2) and the radius is √(9 + 4 + 12) = 5; the point satisfies (6 − 3)² + (2 + 2)² = 25, so it lies on the circle. The radius gradient is (2 + 2)/(6 − 3) = 4/3, so the tangent gradient is −3/4. Hence y − 2 = −(3/4)(x − 6), that is 3x + 4y = 26.
Traps
- Completing the square wrongly and quoting the centre as (g, f) instead of (−g, −f).
- Declaring a line tangent without checking the discriminant or the distance from the centre to the line.
- Discarding the second root when a circle and line intersect twice.