← 返回资讯屏轩教育 · 行业资讯

Chapter 6: Circles — Exercise 6I Review | 第六章:圆 — 练习 6I 复习

2026-08-30 06:50 · 本文部分内容由AI辅助创作

📚 Chapter 6: Circles — Exercise 6I Review | 第六章:圆 — 练习 6I 复习

Exercise 6I is the concluding exercise in Chapter 6 of the AQA A-Level Mathematics Year 1 Pure course. It consolidates every key skill from the chapter: writing equations of circles, finding centres and radii, determining tangents and chords, and solving problems involving intersections between lines and circles. This article gives you a structured, exam-focused revision of exactly what Exercise 6I tests.

练习 6I 是 AQA A-Level 数学 Year 1 纯数部分第六章的收官练习。它综合了本章的所有核心技能:写出圆的方程、求出圆心和半径、确定切线与弦,以及解决直线与圆相交的问题。本文为你提供一份结构化、紧扣考点的复习指南,精准覆盖练习 6I 所考查的内容。


1. The Standard Equation of a Circle | 圆的标准方程

r². In Exercise 6I you must be able to read the centre and radius directly from this form.

圆心为 (a, b)、半径为 r 的圆,其标准方程为 (x − a)² + (y − b)² = r²。当圆心在原点时,方程简化为 x² + y² = r²。在练习 6I 中,你必须能够直接从这种形式读出圆心和半径。

ause r² = 16. Notice that the sign is changed when reading the centre: y + 5 means b = −5.

例如,方程 (x − 3)² + (y + 5)² = 16 表示圆心为 (3, −5)、半径为 4 的圆,因为 r² = 16。注意在读圆心时要变号:y + 5 意味着 b = −5。


2. The General Form and Completing the Square | 圆的一般方程与配方法

o find the centre and radius from this form, you must complete the square in both x and y.

任何圆的方程都可以展开并整理成一般形式 x² + y² + 2gx + 2fy + c = 0。要从这种形式求出圆心和半径,必须分别对 x 和 y 配方。

Consider the equation x² + y² − 6x + 4y − 12 = 0. Group the x terms and the y terms:

考虑方程 x² + y² − 6x + 4y − 12 = 0。将 x 项和 y 项分别分组:

(x² − 6x) + (y² + 4y) − 12 = 0

h simplifies to (x − 3)² + (y + 2)² = 25. Hence the centre is (3, −2) and the radius is 5.

配方得:(x − 3)² − 9 + (y + 2)² − 4 − 12 = 0,化简为 (x − 3)² + (y + 2)² = 25。因此圆心为 (3, −2),半径为 5。


3. Finding the Centre and Radius from the General Form | 从一般式求圆心与半径

For the general form x² + y² + 2gx + 2fy + c = 0, the centre is (−g, −f) and the radius is √g² + f² − c. This formula is useful in Exercise 6I when you are given an expanded equation and need quick answers.

对于一般式 x² + y² + 2gx + 2fy + c = 0,圆心为 (−g, −f),半径为 √g² + f² − c。在练习 6I 中,当你拿到展开后的方程需要快速求解时,这个公式非常有用。

² + 8x − 2y + 1 = 0. Here g = 4, f = −1, c = 1, so the centre is (−4, 1) and the radius is √16 + 1 − 1 = √16 = 4.

以 x² + y² + 8x − 2y + 1 = 0 为例:此处 g = 4, f = −1, c = 1,所以圆心为 (−4, 1),半径为 √16 + 1 − 1 = √16 = 4。

Centre = (−g, −f), Radius = √g² + f² − c

圆心 = (−g, −f),半径 = √g² + f² − c


4. The Tangent–Radius Property | 切线与半径的垂直关系

A tangent to a circle is perpendicular to the radius at the point of contact. This is the most important circle theorem in Exercise 6I. If you know the gradient of the radius, the tangent's gradient is its negative reciprocal.

圆的切线在切点处垂直于半径。这是练习 6I 中最重要的圆定理。如果你知道半径的斜率,切线的斜率就是它的负倒数。

For a circle with centre C and a point P on the circle, the gradient of CP is m, so the tangent at P has gradient −1/m. Always state this relationship before substituting values.

对于圆心为 C、圆上一点 P,CP 的斜率为 m,则 P 处切线的斜率为 −1/m。代入数值前务必先写出这一关系。

o the tangent has gradient −3/4. Its equation is y − 4 = −3/4(x − 3), giving 3x + 4y = 25.

例:圆 x² + y² = 25,求 P(3, 4) 处的切线。半径 OP 的斜率为 4/3,因此切线斜率为 −3/4。切线方程为 y − 4 = −3/4(x − 3),即 3x + 4y = 25。


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

To find the intersection of a line and a circle, substitute the line equation into the circle equation. This gives a quadratic in x (or y). The discriminant then classifies the three possible cases.

要求直线与圆的交点,将直线方程代入圆的方程,得到关于 x(或 y)的二次方程。通过判别式可判断三种情况。

Worked example: Does the line y = x + 7 intersect the circle x² + y² = 25? Substitute y = x + 7 into x² + y² = 25:

例题:直线 y = x + 7 与圆 x² + y² = 25 是否相交?将 y = x + 7 代入 x² + y² = 25:

x² + (x + 7)² = 25 → 2x² + 14x + 24 = 0 → x² + 7x + 12 = 0

e discriminant is Δ = 49 − 48 = 1 > 0, so the line cuts the circle at two distinct points.

判别式 Δ = 49 − 48 = 1 > 0,所以直线与圆有两个不同交点。


6. Chord and Perpendicular Bisector | 弦与垂直平分线

A chord of a circle is a line segment joining two points on the circle. The perpendicular bisector of any chord always passes through the centre of the circle. Exercise 6I often asks you to find the centre using two perpendicular bisectors.

圆的弦是连接圆上两点的线段。任何弦的垂直平分线必定经过圆心。练习 6I 常要求你利用两条垂直平分线求圆心。

. Repeating this with another chord and solving the two bisector equations simultaneously locates the centre.

已知圆上两点 A(1, 2) 和 B(5, 6),AB 的中点为 (3, 4),AB 的斜率为 1,所以垂直平分线的斜率为 −1。其方程为 y − 4 = −(x − 3),即 y = −x + 7。用另一条弦重复此过程,联立两条垂直平分线方程即可求出圆心。


广告 / Sponsored

7. Finding the Equation of a Circle from Three Points | 由三点求圆的方程

itute each point into x² + y² + 2gx + 2fy + c = 0 to produce three simultaneous equations.

如果已知圆上三个点,可以通过求解一般式中的未知数来确定圆的方程。将每个点代入 x² + y² + 2gx + 2fy + c = 0,得到三个联立方程。

Alternatively, find two perpendicular bisectors as described in Section 6 to locate the centre, then use the distance from the centre to any point as the radius. Both methods appear in Exercise 6I and its exam counterparts.

另一种方法是按第 6 节的思路求出两条垂直平分线,从而确定圆心,再用圆心到任意一点的距离作为半径。这两种方法在练习 6I 及其对应考题中都会出现。


8. Worked Example: A Full Exercise 6I-style Question | 完整例题:练习 6I 风格的综合题

A circle has centre C(2, −1) and passes through the point P(5, 3). (a) Find the radius of the circle. (b) Write down the equation of the circle. (c) Find the equation of the tangent at P. (d) Find where the tangent crosses the x-axis.

圆 C 的圆心为 C(2, −1),且经过点 P(5, 3)。(a) 求圆的半径;(b) 写出圆的方程;(c) 求 P 处切线的方程;(d) 求切线与 x 轴的交点。

(a) The radius r = CP = √((5 − 2)² + (3 − (−1))²) = √9 + 16 = √25 = 5.

(a) 半径 r = CP = √((5 − 2)² + (3 − (−1))²) = √9 + 16 = √25 = 5。

(b) The equation is (x − 2)² + (y + 1)² = 25.

(b) 圆的方程为 (x − 2)² + (y + 1)² = 25。

the tangent gradient is −3/4. Thus y − 3 = −3/4(x − 5), which simplifies to 3x + 4y = 27.

(c) CP 的斜率为 (3 − (−1))/(5 − 2) = 4/3,因此切线斜率为 −3/4。所以 y − 3 = −3/4(x − 5),化简得 3x + 4y = 27。

(d) Setting y = 0 gives 3x = 27, so x = 9. The tangent crosses the x-axis at (9, 0).

(d) 令 y = 0,得 3x = 27,即 x = 9。切线与 x 轴交于点 (9, 0)。


9. Common Mistakes and How to Avoid Them | 常见错误及规避方法

2)² which gives a = −2. Always rewrite (x + 2)² as (x − (−2))² before reading the centre.

常见错误之一是从标准方程中读错圆心,尤其是当表达式含加号时,例如 (x + 2)² 对应 a = −2。阅读圆心前,应先将 (x + 2)² 改写成 (x − (−2))²。

Another frequent mistake is forgetting to halve the coefficients when completing the square. For x² − 10x, the square is (x − 5)² − 25, not (x − 10)². Check your completed square by expanding it back.

另一个高频错误是配方时忘记将系数减半。对于 x² − 10x,配方结果是 (x − 5)² − 25,而不是 (x − 10)²。可以通过展开来检验配方是否正确。


10. Exam Tips and Formula Summary | 考点提示与公式总结

In the AQA examination, circle questions typically appear as part of the Pure Mathematics papers, worth between 4 and 8 marks. They test equation writing, coordinate geometry and algebraic substitution in equal measure.

在 AQA 考试中,圆类题目通常出现在纯数试卷中,分值为 4 到 8 分。这类题目均衡地考查方程写法、坐标几何和代数代入能力。

When solving circle problems, always write down the relevant formula before substituting numbers. Show the discriminant explicitly when deciding whether a line intersects a circle. These working lines earn method marks even if the final answer is wrong.

解决圆类问题时,务必先写出相关公式再代入数值。判断直线与圆是否相交时,要明确写出判别式。即使最终答案有误,这些步骤也能获得方法分。

Concept | 概念Formula | 公式
Standard equation | 标准方程(x − a)² + (y − b)² = r²
General form | 一般式x² + y² + 2gx + 2fy + c = 0
Centre | 圆心(−g, −f)
Radius | 半径√g² + f² − c
Tangent condition | 切线条件Discriminant = 0 | 判别式 = 0
Tangent–radius | 切线与半径m_tangent × mradius = −1

Master these formulas and practise substituting carefully. Exercise 6I is designed to prepare you for exactly the style of question that appears in the real AQA papers, so working through it thoroughly is one of the most effective revision strategies available.

掌握这些公式并仔细练习代入计算。练习 6I 的设计目的正是让你熟悉 AQA 真实试卷中的题目风格,因此认真完成该练习是最有效的复习策略之一。


Published by TutorHao | Mathematics Revision Series | aleveler.com

广告 / Sponsored
ALevel国际高中数学课程 剑桥牛津培生 CAIE Edexcel AQA国际考试局 课本练习册复习指南 英文教材进…
数学 备考教辅 · 正版现货 · 与本文内容相关
查看详情 / View on Taobao参考价 ¥69.8

更多咨询请联系16621398022(同微信)

备考资讯每日更新,关注不错过
考纲变化 · 科目分析 · 学习方法
扫码关注公众号 tutorhao
在微信里长按二维码识别关注
或截图后用「扫一扫 → 相册」选择本图
公众号:tutorhao