Mixed Exercise 6: Circles | 第6章综合练习：圆

📚 Mixed Exercise 6: Circles | 第6章综合练习：圆

Mixed Exercise 6 brings together the main skills from the circles chapter: writing circle equations, completing the square, finding intersections, tangents, chords and applying circle theorems in coordinate geometry. This revision guide follows the Edexcel A Level Pure Mathematics specification and models the level of reasoning expected in exam questions.

综合练习 6 汇集了圆一章的主要技能：写出圆的方程、配方法、求交点、切线、弦，以及在坐标几何中应用圆定理。本复习指南遵循 Edexcel A Level 纯数学考试大纲，并示范考试题所要求的推理水平。

1. Standard Circle Equation | 圆的标准方程

You need to move confidently between the centre-radius form and the expanded general form. The standard equation of a circle with centre (a, b) and radius r is:

你需要能够在圆心-半径形式和展开的一般形式之间熟练转换。以 (a, b) 为圆心、r 为半径的圆的标准方程为：

(x – a)² + (y – b)² = r²

The general form is written as:

一般形式写作：

x² + y² + 2gx + 2fy + c = 0

Comparing coefficients shows that the centre is (-g, -f) and the radius is given by √(g² + f² – c), provided that g² + f² – c > 0.

通过比较系数可知圆心为 (-g, -f)，半径为 √(g² + f² – c)，其中要求 g² + f² – c > 0。

2. Completing the Square | 配方法求圆心与半径

A circle equation is often given in expanded form, so completing the square is the key routine skill. For example, to find the centre and radius of x² + y² – 6x + 4y – 3 = 0:

圆的方程经常以展开形式给出，因此配方法是关键的常规技能。例如，要求 x² + y² – 6x + 4y – 3 = 0 的圆心和半径：

x² – 6x + y² + 4y – 3 = 0

(x – 3)² – 9 + (y + 2)² – 4 – 3 = 0

(x – 3)² + (y + 2)² = 16

So the centre is (3, -2) and the radius is √16 = 4. Watch the sign: the centre is (3, -2), not (3, 2), because the equation contains (y + 2)².

因此圆心为 (3, -2)，半径为 √16 = 4。注意符号：圆心是 (3, -2) 而不是 (3, 2)，因为方程中含有 (y + 2)²。

3. Intersections of Lines and Circles | 直线与圆的交点

To find where a line and a circle meet, substitute the line equation into the circle equation to produce a quadratic in one variable. The discriminant Δ = b² – 4ac then tells you how many intersections exist.

要求直线与圆的交点，可将直线方程代入圆的方程，得到关于一个变量的二次方程。判别式 Δ = b² – 4ac 能告诉你交点的个数。

Discriminant condition	Geometric meaning	中文含义
Δ > 0	Two distinct intersection points	两个不同交点
Δ = 0	Tangency: exactly one point	相切：恰好一个点
Δ < 0	No intersection	没有交点

Remember that the discriminant method is usually quicker than solving the full quadratic, especially when a question only asks for the number of intersections.

记住，如果题目只要求交点的个数，使用判别式通常比完全解二次方程更快。

4. Tangent and Chord Properties | 切线与弦的性质

Two geometric facts are used constantly in coordinate geometry questions:

以下两个几何性质在坐标几何题中会经常用到：

A tangent to a circle is perpendicular to the radius at the point of contact.
圆的切线与过切点的半径垂直。
The perpendicular bisector of any chord passes through the centre of the circle.
任意弦的垂直平分线都经过圆心。

These facts allow you to convert gradient conditions and midpoint conditions into equations involving the circle centre and radius.

这些性质能帮助你把斜率条件和线段中点条件转化为涉及圆心和半径的方程。

5. Finding Tangents | 求切线方程

If a point (x₁, y₁) lies on a circle centred at the origin with equation x² + y² = r², the tangent at that point is:

如果点 (x₁, y₁) 在以原点为圆心的圆 x² + y² = r² 上，则该点处的切线方程为：

x x₁ + y y₁ = r²

For a circle centred at (a, b), the tangent at (x₁, y₁) is given by:

对于圆心在 (a, b) 的圆，点 (x₁, y₁) 处的切线方程为：

(x₁ – a)(x – a) + (y₁ – b)(y – b) = r²

If you are given a tangent with a fixed gradient m, write the line as y = mx + c. Then set the perpendicular distance from the centre to the line equal to the radius and solve for c:

如果已知切线的固定斜率 m，可将切线写作 y = mx + c。然后令圆心到直线的垂直距离等于半径，解出 c：

|m a – b + c| / √(m² + 1) = r

Here (a, b) is the centre. This method works because a tangent is exactly one radius length away from the centre along the perpendicular direction.

这里的 (a, b) 是圆心。该方法之所以有效，是因为切线沿垂直方向到圆心的距离恰好为一个半径长度。

6. Circle Theorems in Coordinate Geometry | 坐标几何中的圆定理

The circle theorem ‘the angle in a semicircle is a right angle’ is very useful when a diameter is involved. If A and B are endpoints of a diameter, then for any point P on the circle, the vectors PA and PB are perpendicular, so their dot product is zero:

当题目涉及直径时，圆定理“半圆上的圆周角是直角”非常有用。如果 A 和 B 是直径的两个端点，那么对于圆上任意一点 P，向量 PA 和 PB 互相垂直，因此它们的点积为零：

(x – x₁)(x – x₂) + (y – y₁)(y – y₂) = 0

This gives a quick way to write down the equation of a circle from the endpoints of a diameter.

这提供了一种由直径端点快速写出圆的方程的方法。

For two circles with radii r₁ and r₂ and centre distance d, their relative positions are determined by comparing d with r₁ + r₂ and |r₁ – r₂|. This is closely linked to the discriminant method when solving the two circle equations simultaneously.

对于半径分别为 r₁ 和 r₂、圆心距离为 d 的两个圆，它们的位置关系取决于 d 与 r₁ + r₂ 以及 |r₁ – r₂| 的比较。这与联立两个圆方程时使用判别式的方法密切相关。

7. Length of a Chord | 弦长计算

If you know the perpendicular distance d from the centre of a circle to a chord, and the radius is r, then the chord length L is:

如果已知圆心到一条弦的垂直距离 d，且圆的半径为 r，则弦长 L 为：

L = 2√(r² – d²)

This comes from the fact that the perpendicular from the centre to a chord bisects the chord, creating a right-angled triangle with sides d, L/2 and hypotenuse r.

这个公式来源于：圆心到弦的垂线平分弦，构成一个直角三角形，其两条直角边为 d 和 L/2，斜边为 r。

In coordinate questions, d is found by applying the perpendicular distance formula from the centre to the line containing the chord.

在坐标题中，d 可以通过圆心到弦所在直线的垂直距离公式求得。

8. Circles from Geometric Conditions | 由几何条件求圆的方程

A circle can be determined by three non-collinear points. Substitute the three points into the general form x² + y² + 2gx + 2fy + c = 0 and solve the resulting simultaneous equations for g, f and c.

一个圆可以由三个不共线的点确定。将三个点代入一般式 x² + y² + 2gx + 2fy + c = 0，然后解出关于 g、f、c 的联立方程组即可。

If a diameter has endpoints A and B, the centre is the midpoint of AB and the radius is half of the length AB. The circle equation can then be written immediately as:

如果直径的两个端点为 A 和 B，则圆心是 AB 的中点，半径是 AB 长度的一半。此时圆的方程可以直接写成：

(x – x₁)(x – x₂) + (y – y₁)(y – y₂) = 0

This is an efficient alternative to finding the midpoint and radius separately.

这是比单独求中点和半径更高效的一种替代方法。

9. Exam-Style Worked Examples | 考试风格例题

Example 1: Find the tangent to the circle x² + y² = 25 at the point (3, 4).

例题 1：求圆 x² + y² = 25 在点 (3, 4) 处的切线。

x x₁ + y y₁ = r² → 3x + 4y = 25

So the tangent equation is 3x + 4y = 25. A quick check confirms that (3, 4) lies on both the circle and the tangent.

因此切线方程为 3x + 4y = 25。快速检验可以确认 (3, 4) 同时在圆和切线上。

Example 2: Find the points of intersection of the line y = 2x + 1 with the circle (x – 1)² + (y + 2)² = 20.

例题 2：求直线 y = 2x + 1 与圆 (x – 1)² + (y + 2)² = 20 的交点。

Substitute y = 2x + 1 into the circle:

将 y = 2x + 1 代入圆的方程：

(x – 1)² + (2x + 3)² = 20

x² – 2x + 1 + 4x² + 12x + 9 = 20

5x² + 10x – 10 = 0

x² + 2x – 2 = 0

x = -1 ± √3

The corresponding y-values are y = -1 + 2√3 and y = -1 – 2√3, so the intersection points are:

对应的 y 值分别为 y = -1 + 2√3 和 y = -1 – 2√3，因此交点为：

(-1 + √3, -1 + 2√3) and (-1 – √3, -1 – 2√3)

10. Common Mistakes and Tips | 常见错误与应试技巧

Avoid these common mistakes in Mixed Exercise 6 and similar exam questions:

在综合练习 6 及类似的考试题中，请避免以下常见错误：

Forgetting to take the square root when finding the radius from g² + f² – c.
由 g² + f² – c 求半径时忘记开平方。
Making sign errors when stating the centre from the completed square form.
从配方式写出圆心时出现符号错误。
Using Δ > 0 for a tangent; a tangent always gives Δ = 0.
把相切误用为 Δ > 0；相切始终对应 Δ = 0。
Using the distance from the centre to an arbitrary point on the line instead of the perpendicular distance when applying the tangent condition.
应用相切条件时，使用圆心到直线上任意一点的距离，而不是到直线的垂直距离。
Applying the tangent formula x x₁ + y y₁ = r² before checking that (x₁, y₁) actually lies on the circle.
在未检查 (x₁, y₁) 是否真的在圆上之前，就套用切线公式 x x₁ + y y₁ = r²。

Write down the circle centre and radius first, then decide which method is fastest: completing the square, discriminant, perpendicular distance, or a circle-theorem shortcut.

先写出圆心和半径，然后判断哪种方法最快：配方法、判别式、垂直距离，还是圆定理的简便路径。

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Mixed Exercise 6: Circles | 第6章综合练习：圆