Chapter Review 6: Circles | 第6章复习：圆

📚 Chapter Review 6: Circles | 第6章复习：圆

In Edexcel A-Level Pure Mathematics, Chapter 6 covers circles in the coordinate plane. You are expected to find the centre and radius, write equations in standard and general form, and solve problems involving tangents, chords, and intersections with lines or other circles.

在 Edexcel A-Level 纯数学中，第6章研究坐标平面中的圆。你需要掌握求圆心和半径、写出标准式和一般式，以及解决涉及切线、弦以及与直线或另一个圆相交的问题。

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

The equation of a circle with centre (a, b) and radius r is:

圆心在 (a, b)、半径为 r 的圆的方程是：

(x – a)² + (y – b)² = r²

If the circle is centred at the origin, this simplifies to x² + y² = r². The radius must be positive, and the right-hand side is r², so the left-hand side cannot be negative for points on the circle.

如果圆心在原点，方程简化为 x² + y² = r²。半径必须为正数，右边是 r²，因此圆上的点满足左边不能为负。

You should be able to write the standard equation directly when given the centre and radius. For example, a circle with centre (-2, 5) and radius 4 has equation (x + 2)² + (y – 5)² = 16.

你应该能够根据给出的圆心和半径直接写出标准方程。例如，圆心为 (-2, 5)、半径为 4 的圆，其方程为 (x + 2)² + (y – 5)² = 16。

2. General Form and Identifying Centre and Radius | 一般式与识别圆心和半径

The expanded general form of a circle is:

圆的一般式展开为：

x² + y² + 2gx + 2fy + c = 0

Comparing with the standard form (x – a)² + (y – b)² = r² gives centre (-g, -f) and radius √(g² + f² – c). The expression g² + f² – c must be positive for a real circle; if it equals zero, the circle degenerates to a point.

与标准式 (x – a)² + (y – b)² = r² 比较，可得圆心为 (-g, -f)，半径为 √(g² + f² – c)。要使圆为实圆，g² + f² – c 必须大于零；如果等于零，该圆退化成一个点。

Example: x² + y² – 6x + 8y – 11 = 0 has centre (3, -4) and radius √(9 + 16 + 11) = √36 = 6.

例如：x² + y² – 6x + 8y – 11 = 0 的圆心为 (3, -4)，半径为 √(9 + 16 + 11) = √36 = 6。

3. Completing the Square | 配方法

To convert a general circle equation into standard form, complete the square for the x terms and the y terms separately. This is the most reliable way to find the centre and radius from a general form.

要把一般式转化为标准式，需要对 x 项和 y 项分别配方。这是从一般式求圆心和半径最可靠的方法。

Example: x² + y² + 10x – 4y + 20 = 0 becomes (x + 5)² – 25 + (y – 2)² – 4 + 20 = 0, so (x + 5)² + (y – 2)² = 9, giving centre (-5, 2) and radius 3.

例如：x² + y² + 10x – 4y + 20 = 0 变为 (x + 5)² – 25 + (y – 2)² – 4 + 20 = 0，因此 (x + 5)² + (y – 2)² = 9，圆心为 (-5, 2)，半径为 3。

Be careful: the constant term moves to the right-hand side, and you must add the same values within the brackets to both sides of the equation to keep it balanced.

注意：常数项移到右边，配方时在括号内加上的值必须同时加到方程两边以保持平衡。

4. Intersections with a Line | 与直线的交点

To find where a line intersects a circle, substitute the line equation y = mx + c into the circle equation. This produces a quadratic in x, which can be solved using factorisation, the quadratic formula, or the discriminant.

要求直线与圆的交点，将直线方程 y = mx + c 代入圆的方程，得到关于 x 的一元二次方程。可以使用因式分解、求根公式或判别式求解。

If the discriminant is positive, there are two intersections; if it is zero, the line is a tangent; if it is negative, the line misses the circle. This discriminant test is often used to show tangency.

如果判别式大于零，则有两个交点；如果等于零，直线为切线；如果小于零，直线与圆不相交。这个判别式检验常用于证明直线与圆相切。

Example: Determine whether y = x + 1 intersects x² + y² = 5. Substitute: x² + (x + 1)² = 5 → 2x² + 2x – 4 = 0 → x² + x – 2 = 0, so x = 1 or x = -2, giving two points of intersection.

例如：判断 y = x + 1 是否与 x² + y² = 5 相交。代入得 x² + (x + 1)² = 5 → 2x² + 2x – 4 = 0 → x² + x – 2 = 0，所以 x = 1 或 x = -2，得到两个交点。

5. Tangents to a Circle | 圆的切线

A tangent to a circle touches the circle at exactly one point. The tangent is perpendicular to the radius at the point of contact. This perpendicular relationship is central to most tangent questions.

圆的切线在圆上恰好接触一点。切线在切点处垂直于半径。这个垂直关系是大多数切线问题的核心。

If the gradient of the radius to point P is m_r, then the gradient of the tangent at P is m_t = -1/m_r, provided m_r ≠ 0. If the radius is vertical, the tangent is horizontal.

如果半径到点 P 的斜率为 m_r，那么 P 点切线的斜率为 m_t = -1/m_r，前提是 m_r ≠ 0。如果半径是竖直的，切线就是水平的。

To find the tangent equation, use the point-slope form y – y₁ = m_t(x – x₁). You may also need to prove a line is tangent by showing the perpendicular distance from the centre equals the radius, or by showing the discriminant is zero.

求切线方程时，使用点斜式 y – y₁ = m_t(x – x₁)。你可能还需要通过证明圆心到直线的垂直距离等于半径，或证明判别式为零，来证明一条直线是切线。

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

A chord is a line segment joining two points on a circle. The perpendicular bisector of any chord passes through the centre of the circle. This property is extremely useful for locating the centre when only a chord is known.

弦是圆上两点之间的线段。任何弦的垂直平分线都经过圆心。这个性质在已知一条弦而求圆心时非常有用。

If you know the midpoint of a chord and its gradient, the line through the centre is perpendicular to the chord. You can use this to find the centre or to write an equation the centre must satisfy.

如果已知弦的中点及其斜率，经过圆心的直线与该弦垂直。你可以用这个条件求出圆心，或者写出圆心必须满足的方程。

The shortest distance from the centre to a chord is the perpendicular distance, and this distance d, the chord length L, and the radius r are related by r² = d² + (L/2)². This is Pythagoras applied to half of the chord.

圆心到弦的最短距离是垂直距离，该距离 d、弦长 L 和半径 r 满足 r² = d² + (L/2)²。这就是把勾股定理应用于半弦得到的。

7. Equation of a Chord and Circle Problems | 弦与圆问题

To find the equation of a chord given two points on the circle, use the midpoint and gradient formula. Alternatively, if the chord is defined by an intersection, solve the line and circle equations simultaneously.

如果已知圆上两点，求弦的方程可以使用中点和斜率公式。如果弦由交点定义，则联立直线与圆的方程求解。

Example: The chord through A(1, 2) and B(5, 6) has midpoint (3, 4) and gradient (6 – 2)/(5 – 1) = 1. Its perpendicular bisector has gradient -1 and passes through (3, 4), so its equation is y – 4 = -1(x – 3).

例如：经过 A(1, 2) 和 B(5, 6) 的弦，其中点为 (3, 4)，斜率为 (6 – 2)/(5 – 1) = 1。其垂直平分线斜率为 -1，且经过 (3, 4)，所以方程为 y – 4 = -1(x – 3)。

If this chord belongs to a circle, the centre must lie on the line y = -x + 7. This is often the key step in finding the circle’s equation from a chord and one other condition.

如果这条弦属于某个圆，圆心必定在直线 y = -x + 7 上。这往往是已知一条弦和另一个条件求圆的方程时的关键步骤。

8. Intersection of Two Circles | 两圆的交点

To find the common points of two circles, subtract one equation from the other to eliminate the x² and y² terms. This gives a linear equation, which often represents the common chord, also called the radical axis.

要求两圆的公共交点，将两个方程相减以消去 x² 和 y² 项，得到一条直线方程，这条直线通常表示公共弦，也叫根轴。

Then solve this line with one of the circle equations to find the intersection points, if any exist. If the distance between centres is greater than the sum of radii, the circles do not intersect.

然后将该直线与其中一个圆的方程联立求解，得到交点（如果有）。如果圆心距大于半径之和，两个圆不相交。

Example: Circle C₁: x² + y² = 25 and C₂: (x – 4)² + y² = 9. Subtract: (x – 4)² – x² = -16 → -8x + 16 = -16 → x = 4. Then y² = 25 – 16 = 9, so points (4, 3) and (4, -3).

例如：圆 C₁: x² + y² = 25 与 C₂: (x – 4)² + y² = 9 相减：(x – 4)² – x² = -16 → -8x + 16 = -16 → x = 4。然后 y² =

Published by TutorHao | A-Level Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Chapter Review 6: Circles | 第6章复习：圆