AQA A-level Mathematics 7357, section C (Coordinate geometry), version 1.3 (first assessment 2018). Original teaching notes; objective-level mapping lives in content/teaching/aqa_a_level_mathematics/objective-review.yaml.
Prerequisites: GCSE straight lines, Pythagoras and quadratic equations. Each lesson develops its own method before independent transfer.
These are original teaching examples, not official AQA material.
3.2
Coordinate geometry 圆
What would explain this observation?
A satellite dish's cross-section and a wheel's rim are both circles, but only the rim can be written (x − a)² + (y − b)² = r² with a centre you can read at a glance. What does that equation buy you?
Start with a prediction. State what the parameters a, b and r each control, and how a tangent direction could be found without differentiation.
Build the model
A line through (x₁, y₁) with gradient 斜率 m is y − y₁ = m(x − x₁). Parallel lines share m; perpendicular finite gradients satisfy m₁ × m₂ = −1.
A circle with centre (a, b) and radius r is (x − a)² + (y − b)² = r². The radius to a contact point is perpendicular to the tangent 切线 there, so the tangent's gradient is the negative reciprocal of the radius's. A diameter subtends a right angle at the circumference.
gradient: The change in y per unit change in x; circle: The set of points at a fixed distance from a centre; tangent: A line touching the circle at one point, perpendicular to the radius there.
Choose evidence that can test it
Check that a proposed point lies on the circle by substituting into (x − a)² + (y − b)² = r² before building a tangent at it. Complete the square to rescue a centre from the expanded form x² + y² + px + qy + s = 0 — and adjust the right side too.
Intersect a line with the circle by substitution: a repeated root means tangency, two roots a chord, and a negative discriminant no meeting.
Work from known quantities
State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
Known: the circle x² + y² = 25 and the point (3, 4) on it (3² + 4² = 25 ✓). The radius from (0, 0) to (3, 4) has gradient 4/3, so the tangent gradient is −3/4 by m₁ × m₂ = −1. The tangent is y − 4 = −(3/4)(x − 3), which rearranges to 3x + 4y = 25.
Example:
Find the centre and radius of x² + y² − 4x + 2y − 20 = 0 using the same sequence: group terms → complete each square → read the form. (Answer: (x − 2)² + (y + 1)² = 25, so the centre is (2, −1) and the radius is 5; note the bracket signs reverse in the centre.)
Check the conclusion and its limits
The negative-reciprocal rule needs finite, nonzero gradients; a horizontal radius gives a vertical tangent, stated as x = constant. A chord bisected by a centre-perpendicular is a right-triangle shortcut: half-chord, radius and centre distance obey Pythagoras.
Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.
Warn:
The tangent at a circle point has the same gradient as the radius to that point. This claim is false: the radius and tangent are perpendicular at the contact point, so their gradients multiply to −1 — at (3, 4), the radius has 4/3 and the tangent has −3/4.
Key:
Coordinate geometry: lines by point-gradient, circles by (x − a)² + (y − b)² = r². Tangents use the radius-perpendicular relation; intersections use substitution and the discriminant.
