📚 Mixed Exercise 6: Circles | 混合练习6：圆
Mixed Exercise 6 in the Edexcel AS/A Level Pure Mathematics textbook consolidates all key concepts from Chapter 6: Circles. This exercise challenges students to apply circle equations, tangent and chord properties, and discriminant analysis to solve a variety of problems. Mastering this mixed exercise is essential for success in the exam, as circles frequently appear alongside straight lines and quadratic theory.
Edexcel AS/A Level 纯数学教材中的混合练习6整合了第6章“圆”的所有核心概念。该练习要求学生运用圆的方程、切线和弦的性质以及判别式分析来解决各种问题。掌握该混合练习对考试成功至关重要，因为圆经常与直线和二次理论结合考查。
1. Reviewing the Standard Equation | 回顾标准方程
In coordinate geometry, a circle with centre (a, b) and radius r is described by (x − a)² + (y − b)² = r². If the centre is at the origin, the equation simplifies to x² + y² = r².
在解析几何中，圆心为 (a, b)、半径为 r 的圆由方程 (x − a)² + (y − b)² = r² 表示。如果圆心在原点，方程简化为 x² + y² = r²。
You can also encounter the expanded form x² + y² + 2gx + 2fy + c = 0, where the centre is (−g, −f) and the radius is √(g² + f² − c). This form is often used in exam questions.
你还会遇到展开形式 x² + y² + 2gx + 2fy + c = 0，其圆心为 (−g, −f)，半径为 √(g² + f² − c)。这种形式经常出现在考试题中。
2. Finding Centre and Radius by Completing the Square | 通过配方法求圆心与半径
When a circle’s equation is given in expanded form, you must complete the square for both x and y terms to rewrite it in standard form. This reveals the centre coordinates and the radius directly.
当圆的方程以展开形式给出时，必须对 x 项和 y 项分别完成平方，将其改写为标准形式，从而直接得到圆心坐标和半径。
Example: x² + y² − 6x + 4y − 3 = 0 → (x − 3)² + (y + 2)² = 16 → Centre (3, −2), radius 4.
例子：x² + y² − 6x + 4y − 3 = 0 → (x − 3)² + (y + 2)² = 16 → 圆心 (3, −2)，半径 4。
3. Position of a Point Relative to a Circle | 点与圆的位置关系
To check whether a point lies inside, on, or outside a circle, substitute its coordinates into the left-hand side of the circle equation (x − a)² + (y − b)² and compare the result with r².
要判断一点在圆内、圆上还是圆外，只需将其坐标代入圆的方程 (x − a)² + (y − b)² 的左边，并将结果与 r² 进行比较。
- If (x₁ − a)² + (y₁ − b)² < r², the point is inside the circle.
- 若 (x₁ − a)² + (y₁ − b)² < r²，点在圆内。
- If (x₁ − a)² + (y₁ − b)² = r², the point lies on the circle.
- 若 (x₁ − a)² + (y₁ − b)² = r²，点在圆上。
- If (x₁ − a)² + (y₁ − b)² > r², the point is outside the circle.
- 若 (x₁ − a)² + (y₁ − b)² > r²，点在圆外。
4. Finding the Circle Equation Given a Diameter | 已知直径端点求圆的方程
The midpoint of a diameter gives the centre of the circle. The radius is half the distance between the endpoints, or the distance from the centre to either endpoint. Use the centre and radius to write the equation.
直径两端点的中点即为圆心。半径为两端点距离的一半，或者圆心到任一端点的距离。利用圆心和半径即可写出方程。
Typical mixed exercise question: “The points A(2, 1) and B(8, 9) are the ends of a diameter. Find the equation of the circle.” Midpoint is (5, 5), radius is ½ × distance AB = ½ × 10 = 5, so (x − 5)² + (y − 5)² = 25.
混合练习中的典型题目：“点 A(2, 1) 和 B(8, 9) 是一条直径的端点。求圆的方程。”中点为 (5, 5)，半径为 ½ × AB 距离 = ½ × 10 = 5，因此 (x − 5)² + (y − 5)² = 25。
5. Tangent Properties: Radius Perpendicular to Tangent | 切线性质：半径垂直于切线
A key theorem states that the radius drawn to the point of tangency is perpendicular to the tangent line. This relationship allows you to find the gradient of the tangent from the gradient of the radius.
一个重要定理指出，画出到切点的半径垂直于切线。利用这一关系，可以从半径的斜率求出切线的斜率。
If the centre is C(a, b) and the point of contact is P(x₁, y₁), the gradient of CP is (y₁ − b)/(x₁ − a). The gradient of the tangent is the negative reciprocal: m_tangent = −(x₁ − a)/(y₁ − b) (provided y₁ ≠ b).
若圆心为 C(a, b)，切点为 P(x₁, y₁)，则 CP 的斜率为 (y₁ − b)/(x₁ − a)。切线的斜率为负倒数：m_切线 = −(x₁ − a)/(y₁ − b)（只要 y₁ ≠ b）。
6. Finding the Equation of a Tangent | 求切线方程
Using the perpendicular gradient and the coordinates of the point of contact, apply the point–gradient form y − y₁ = m(x − x₁). Always state the final equation in a tidy linear form ax + by + c = 0 or y = mx + c.
利用垂直斜率和切点坐标，应用点斜式 y − y₁ = m(x − x₁)。最终方程应整理成简洁的线性形式 ax + by + c = 0 或 y = mx + c。
Example: Find the tangent to x² + y² = 25 at the point (3, 4). The centre is (0, 0), radius gradient = 4/3, so tangent gradient = −3/4. Equation: y − 4 = −3/4 (x − 3) → 4y − 16 = −3x + 9 → 3x + 4y = 25.
示例：求圆 x² + y² = 25 在点 (3, 4) 处的切线。圆心为 (0, 0)，半径斜率为 4/3，因此切线斜率为 −3/4。方程：y − 4 = −3/4 (x − 3) → 4y − 16 = −3x + 9 → 3x + 4y = 25。
7. Chord Properties: Perpendicular Bisector Passes Through the Centre | 弦的性质：垂直平分线过圆心
The perpendicular bisector of any chord of a circle passes through the centre. This property is useful for finding the centre when given a chord and a tangent, or for proving a point is the centre.
圆内任意弦的垂直平分线经过圆心。当已知一条弦和一条切线时，这一性质有助于求出圆心，或用于证明某点为圆心。
To find the centre, you can find the equation of the perpendicular bisector of a chord, and then intersect it with another geometrical constraint such as a line through a point of tangency.
要找出圆心，可以先求出弦的垂直平分线方程，然后将其与另一几何条件（如经过切点的直线）联立求解。
8. Intersection of a Line and a Circle: Discriminant Approach | 直线与圆的交点：判别式法
To find the intersection points of a line y = mx + c and a circle (x − a)² + (y − b)² = r², substitute the line equation into the circle equation to form a quadratic in x (or y). Then solve or analyse the discriminant Δ = b² − 4ac.
要求直线 y = mx + c 与圆 (x − a)² + (y − b)² = r² 的交点，可将直线方程代入圆的方程，得到关于 x（或 y）的一元二次方程。然后求解或分析判别式 Δ = b² − 4ac。
- Δ > 0 → two distinct intersection points (line cuts the circle).
- Δ > 0 → 两个相异的交点（直线与圆相交）。
- Δ = 0 → line is tangent to the circle (one repeated point).
- Δ = 0 → 直线与圆相切（一个重合点）。
- Δ < 0 → no intersection (line misses the circle).
- Δ < 0 → 没有交点（直线与圆相离）。
This method is heavily tested in mixed exercise 6, often requiring you to find unknown constants for tangency.
混合练习6 中大量考查此方法，通常需要你求出使直线与圆相切的未知常数。
9. Tangency Condition and Unknown Coefficients | 相切条件与未知系数
A common question gives a circle equation and a line containing an unknown parameter (e.g., y = mx or y = 2x + k), and asks for the value(s) that make the line tangent. Set Δ = 0 and solve for the parameter.
常见题型给出圆的方程和含未知参数的直线（如 y = mx 或 y = 2x + k），要求求出使直线与圆相切的值。令 Δ = 0 并求解参数即可。
For instance, find k such that the line y = 2x + k is tangent to x² + y² = 5. Substitute to get x² + (2x + k)² = 5 → 5x² + 4kx + (k² − 5) = 0. For tangency, discriminant = 16k² − 20(k² − 5) = 0 → −4k² + 100 = 0 → k² = 25 → k = ±5.
例如，求使直线 y = 2x + k 与圆 x² + y² = 5 相切的 k 值。代入得到 5x² + 4kx + (k² − 5) = 0。由相切所以判别式 = 16k² − 20(k² − 5) = 0 → −4k² + 100 = 0 → k² = 25 → k = ±5。
10. Circles and Circles: Intersections and Relative Position | 圆与圆：交点与位置关系
Two circles may intersect at two points, touch externally or internally, or have no intersection. The distance d between centres and the radii r₁, r₂ determine the relationship.
两圆可能相交于两点、外切、内切或相离。圆心距 d 和半径 r₁、r₂ 决定了位置关系。
| Condition | 条件 | Geometry | 几何意义 |
|---|---|
| d > r₁ + r₂ | Circles are externally separate; no intersection. | 两圆外离，无交点。 |
| d = r₁ + r₂ | Touch externally; one common point. | 外切，一个公共点。 |
| |r₁ − r₂| < d < r₁ + r₂ | Intersect at two points. | 相交于两点。 |
| d = |r₁ − r₂| | Touch internally. | 内切。 |
| d < |r₁ − r₂| | One circle lies inside the other, no intersection. | 内含，无交点。 |
Exam questions may ask to find the intersection points algebraically by subtracting equations or to deduce tangency conditions between circles.
考试题可能要求通过方程相减来代数求交点，或推导两圆相切的条件。
11. Applying Circles to Coordinate Geometry Problems |
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