Coordinate geometry applies algebra to geometric problems. Lines, circles, and their intersections are studied using equations — connecting algebraic manipulation with geometric interpretation.
Gradient m = (y₂-y₁)/(x₂-x₁). Parallel lines: m₁ = m₂. Perpendicular lines: m₁ × m₂ = -1. Equation forms: y - y₁ = m(x - x₁), y = mx + c, ax + by + c = 0. Midpoint: ((x₁+x₂)/2, (y₁+y₂)/2). Distance: √((x₂-x₁)² + (y₂-y₁)²). Perpendicular bisector: passes through midpoint, gradient = -1/m.
Standard form: (x-a)² + (y-b)² = r² — centre (a,b), radius r. Expanded form: x² + y² + 2gx + 2fy + c = 0 — centre (-g,-f), radius √(g²+f²-c). Circle through three points: substitute each into general equation, solve simultaneously. Line-circle intersection: substitute line equation into circle equation, solve resulting quadratic — discriminant determines: 2 points (Δ>0), tangent (Δ=0), no intersection (Δ<0).
Tangent to circle at point P: perpendicular to radius at P. Normal at P: along the radius. Equation of tangent: use gradient of radius, then perpendicular gradient. Angle in semicircle = 90° — useful for proving right angles. Perpendicular from centre to chord bisects the chord. Locus problems: set of points satisfying a condition.
Method: (1) Find the gradient of the radius from centre to the point of tangency. (2) The tangent is perpendicular to the radius, so its gradient = -1/(gradient of radius). (3) Use y - y₁ = m(x - x₁) with the tangent gradient and the point. For example: circle centre (2, 3), tangent at point (5, 7). Gradient of radius = (7-3)/(5-2) = 4/3. Gradient of tangent = -3/4. Tangent equation: y - 7 = -3/4(x - 5). This method works for any circle, not just those centred at the origin.
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