Coordinate Geometry for AQA A-Level Maths | A-Level AQA 数学：坐标几何考点精讲

📚 Coordinate Geometry for AQA A-Level Maths | A-Level AQA 数学：坐标几何考点精讲

Coordinate geometry is a central topic in AQA A-Level Mathematics, bridging algebraic techniques with geometric intuition. It covers straight lines, circles, distances, tangents, and parametric equations. Mastery of these ideas is essential for problem-solving across pure mathematics and applied contexts. This article breaks down every key concept, providing clear explanations and worked examples tailored to the AQA specification.

坐标几何是 AQA A-Level 数学的核心主题，它将代数技巧与几何直觉联系起来。内容包括直线、圆、距离、切线以及参数方程。掌握这些概念对于纯数学及其应用领域的解题至关重要。本文逐一解析每个关键考点，提供清晰解释和贴合 AQA 考试要求的例题讲解。

1. Equations of a Straight Line | 直线方程的各种形式

The gradient–intercept form y = mx + c directly shows the slope m and the y‑intercept c. It is the most common representation.

斜截式 y = mx + c 直接给出斜率 m 和 y 轴截距 c，是最常用的表示方法。

The point–slope form y − y₁ = m(x − x₁) is useful when a point (x₁, y₁) and the gradient are known.

点斜式 y − y₁ = m(x − x₁) 在已知一点 (x₁, y₁) 及斜率时非常方便。

The general form ax + by + c = 0 is preferred for vector work and distance of a point from a line.

一般式 ax + by + c = 0 常用于向量问题及求解点到直线的距离。

2. Gradient, Parallel and Perpendicular Lines | 斜率与平行、垂直关系

The gradient between two points (x₁, y₁) and (x₂, y₂) is m = (y₂ − y₁) / (x₂ − x₁). Parallel lines share the same gradient.

两点 (x₁, y₁) 与 (x₂, y₂) 间的斜率为 m = (y₂ − y₁) / (x₂ − x₁)。平行直线斜率相等。

For perpendicular lines, gradients multiply to −1: m₁ × m₂ = −1, provided neither is vertical.

垂直直线的斜率乘积为 −1：m₁ × m₂ = −1，前提是两条线均不竖直。

If a line is horizontal, its gradient is 0; a vertical line has an undefined gradient.

水平线的斜率为 0；竖直线的斜率不存在。

3. Distance Between Two Points | 两点间的距离公式

The distance d between (x₁, y₁) and (x₂, y₂) derives from Pythagoras’ theorem:

两点 (x₁, y₁) 与 (x₂, y₂) 间的距离 d 来自勾股定理：

d = √[(x₂ − x₁)² + (y₂ − y₁)²]

This formula is fundamental for finding lengths of line segments, radii of circles, and distances in locus problems.

该公式是求线段长度、圆的半径以及轨迹问题中距离的基础。

4. Midpoint of a Line Segment | 线段的中点公式

The midpoint M of the segment joining (x₁, y₁) and (x₂, y₂) is given by:

连接 (x₁, y₁) 与 (x₂, y₂) 的线段中点 M 为：

M = ( (x₁ + x₂)/2 , (y₁ + y₂)/2 )

Midpoints are used in geometric proofs, finding perpendicular bisectors, and locating centres of circles.

中点常用于几何证明、求垂直平分线以及确定圆心。

5. Intersection of Two Lines | 两条直线的交点

To find the intersection of two lines, solve their equations simultaneously. For example, given y = 2x − 1 and y = −x + 5, equate the y‑values:

要求两条直线的交点，需联立解它们的方程。例如，已知 y = 2x − 1 和 y = −x + 5，令 y 值相等：

2x − 1 = −x + 5 ⇒ 3x = 6 ⇒ x = 2, y = 3

If using the general form, apply substitution or elimination. The intersection point (2, 3) is often used to check consistency of three or more lines.

若使用一般式，则采用代入法或消元法。交点 (2, 3) 常用来检验三条及以上直线的共点性。

6. Equation of a Circle | 圆的方程

A circle with centre (a, b) and radius r has the standard form:

以 (a, b) 为圆心、r 为半径的圆，标准方程为：

(x − a)² + (y − b)² = r²

Expanding gives the general form x² + y² + 2gx + 2fy + c = 0, where the centre is (−g, −f) and radius r = √(g² + f² − c), provided g² + f² − c > 0.

展开后得到一般式 x² + y² + 2gx + 2fy + c = 0，圆心为 (−g, −f)，半径 r = √(g² + f² − c)，要求 g² + f² − c > 0。

Always complete the square to identify centre and radius from the general equation.

从一般式求圆心和半径时，始终要通过配方法。

7. Position of a Point Relative to a Circle | 点与圆的位置关系

Substitute the point’s coordinates into the circle equation. For a circle (x − a)² + (y − b)² = r²:

将点的坐标代入圆的方程。对于圆 (x − a)² + (y − b)² = r²：

If (x − a)² + (y − b)² < r², the point lies inside.

若 (x − a)² + (y − b)² < r²，点在圆内。
If (x − a)² + (y − b)² = r², the point is on the circumference.

若 (x − a)² + (y − b)² = r²，点在圆上。
If (x − a)² + (y − b)² > r², the point lies outside.

若 (x − a)² + (y − b)² > r²，点在圆外。

This test is essential when finding tangents or intersections with lines.

在求切线或与直线的交点时，这一判断是基础。

8. Intersection of a Line and a Circle | 直线与圆的交点

To find where a line y = mx + c meets a circle, substitute the line equation into the circle equation and solve the resulting quadratic in x.

求直线 y = mx + c 与圆的交点时，将直线方程代入圆的方程，解出关于 x 的二次方程。

The discriminant Δ of the quadratic tells the number of intersections:

该二次方程的判别式 Δ 决定交点个数：

Δ > 0: two distinct intersection points (secant).

Δ > 0：两个不同交点（割线）。
Δ = 0: exactly one point (tangent).

Δ = 0：一个交点（切线）。
Δ < 0: no intersection (line misses the circle).

Δ < 0：无交点（直线与圆相离）。

This discriminant condition is the algebraic foundation for finding tangents.

判别式条件是求切线问题的代数基础。

9. Tangents to a Circle | 圆的切线方程

A tangent to a circle at a point P on the circumference is perpendicular to the radius at P. To find its equation, calculate the gradient of the radius CP, then use the negative reciprocal for the tangent’s gradient.

圆上一点 P 处的切线与该点到圆心的半径垂直。求切线方程时，先求半径 CP 的斜率，再取负倒数作为切线斜率。

If the tangent is drawn from an external point, set the discriminant of the line–circle quadratic to zero (Δ = 0) and solve for the unknown gradient m.

若切线从圆外一点引出，则令直线与圆的二次方程判别式为零（Δ = 0），解出未知斜率 m。

The point of tangency can then be found by solving the equal‑slope condition or by using the fact that the distance from the centre to the line equals the radius.

然后可通过斜率相等条件或圆心到直线的距离等于半径来求切点。

10. Perpendicular Bisector of a Chord | 弦的垂直平分线

A fundamental property: the perpendicular from the centre of a circle to a chord bisects the chord. Conversely, the line joining the centre to the midpoint of a chord is perpendicular to the chord.

一个基本性质是：从圆心到弦的垂线平分该弦。反之，连接圆心与弦中点的直线垂直于该弦。

To find the perpendicular bisector of a chord, first locate the midpoint of the chord, then compute the gradient perpendicular to the chord, and write the line equation. This line always passes through the centre of the circle.

求弦的垂直平分线时，先找到弦的中点，然后求出垂直于弦的斜率，写出直线方程。这条垂直平分线必过圆心。

This property is extremely useful for determining the centre of a circle when only three points on the circumference are known.

当只给出圆上三点时，这一性质对确定圆心极为有用。

11. Parametric Equations of a Line and a Circle | 直线与圆的参数方程

A straight line through (x₀, y₀) with direction vector (d₁, d₂) can be written in parametric form:

过点 (x₀, y₀) 且方向向量为 (d₁, d₂) 的直线可写成参数形式：

x = x₀ + t d₁, y = y₀ + t d₂

A circle centre (a, b) radius r has the parametric equations:

以 (a, b) 为圆心、r 为半径的圆，参数方程为：

x = a + r cos θ, y = b + r sin θ

Parametric forms simplify problems involving motion, intersection with other curves, and differentiation (e.g. finding gradients).

参数形式可简化涉及运动、与其他曲线相交以及求导（如斜率）的问题。

12. Worked Application Example | 综合应用示例

Find the centre and radius of the circle passing through the points A(1, 2), B(3, −4) and C(5, 6).

求过点 A(1, 2)、B(3, −4) 和 C(5, 6) 的圆的圆心和半径。

Step 1: Find the perpendicular bisectors of two chords, say AB and BC. Midpoint of AB is (2, −1); gradient AB is −3, so perpendicular gradient = 1/3. Equation of perpendicular bisector: y + 1 = (1/3)(x − 2).

步骤1：求两条弦（如 AB 和 BC）的垂直平分线。AB 中点为 (2, −1)；AB 斜率为 −3，因此垂直平分线斜率为 1/3。方程为 y + 1 = (1/3)(x − 2)。

Midpoint of BC is (4, 1); gradient BC is 5, perpendicular gradient = −1/5. Equation: y − 1 = −(1/5)(x − 4).

BC 中点为 (4, 1)；BC 斜率为 5，垂直平分线斜率为 −1/5。方程：y − 1 = −(1/5)(x − 4)。

Step 2: Solve the two linear equations simultaneously to find the centre O. The intersection gives O(3, 0).

步骤2：联立解这两条直线方程求圆心 O。解得交点为 O(3, 0)。

Step 3: Radius r = distance OA = √[(3 − 1)² + (0 − 2)²] = √(4 + 4) = √8 = 2√2. Circle equation: (x − 3)² + y² = 8.

步骤3：半径 r = OA 的距离 = √[(3 − 1)² + (0 − 2)²] = √8 = 2√2。圆的方程为 (x − 3)² + y² = 8。

This method blends midpoints, perpendicular gradients, and distance calculations—a classic AQA exam problem.

该方法综合了中点、垂直斜率及距离计算——是 AQA 考试的经典题型。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Coordinate Geometry for AQA A-Level Maths | A-Level AQA 数学：坐标几何 考点精讲