Edexcel A-Level数学圆的方程核心考点与解法 | Edexcel A-Level Maths: Circle Equations Key Points and Solutions

Edexcel A-Level数学圆的方程核心考点与解法

一、圆的方程两种标准形式 | Two Standard Forms of a Circle Equation

Edexcel A-Level数学纯数第一章到第六章中，圆（Circles）这一章虽然内容不长，却是考试中高频出现的基础章节。圆的方程有两种标准形式：圆心在(a, b)、半径为r的圆，方程为(x-a)^2 + (y-b)^2 = r^2；展开后得到一般式x^2 + y^2 + 2gx + 2fy + c = 0，其中圆心为(-g, -f)，半径为根号下(g^2 + f^2 - c)。考试中很多题目会先给出一般式，要求你通过配方（completing the square）还原成标准式，从而直接读出圆心和半径。掌握这两种形式的互相转换，是做对本章所有大题的第一步，也是最容易拿分的一步。

There are two standard forms for the equation of a circle in Edexcel A-Level Pure Mathematics. A circle with centre (a, b) and radius r is written as (x-a)^2 + (y-b)^2 = r^2. Expanding the brackets produces the general form x^2 + y^2 + 2gx + 2fy + c = 0, where the centre is (-g, -f) and the radius is the square root of (g^2 + f^2 - c). In the exam, many questions give the general form first and ask you to complete the square to recover the standard form, so that the centre and radius can be read directly. Mastering this conversion is the first step toward full marks on this chapter, and it is also the easiest mark to secure.

二、由圆心与半径直接写方程 | Writing the Equation from Centre and Radius

最直接的题型是已知圆心坐标和半径，直接写出圆的方程。例如圆心为(5, -3)、半径为4的圆，直接代入标准式得到(x-5)^2 + (y+3)^2 = 16。注意y坐标是-3时，标准式里写成y+3，符号最容易出错。课本第135页混合练习中，第10题给出圆心在(5, 1)、点A(-3, -7)在圆上，这时半径等于圆心到点A的距离，用两点间距离公式求出半径，再写方程。这类题考察的是对标准式结构的理解，只要圆心和半径找对，代入过程几乎不丢分。

The simplest question type gives the centre and radius directly and asks for the equation. For example, a circle with centre (5, -3) and radius 4 is written as (x-5)^2 + (y+3)^2 = 16. Notice that when the y-coordinate is -3, the standard form contains y+3, and sign errors here are extremely common. In the mixed exercise on page 135, question 14 gives a circle with centre (5, 1) and tells you that point A(-3, -7) lies on the circle; the radius is therefore the distance between the centre and A, found with the distance formula, before writing the equation. This type tests your understanding of the standard form structure, and once the centre and radius are correct, substitution rarely loses marks.

三、一般式与标准式的换算 | Converting Between General and Standard Form

当题目给出x^2 + y^2 + 6x - 2y = 7这样的方程时，第一步永远是配方：x^2 + 6x配成(x+3)^2 - 9，y^2 - 2y配成(y-1)^2 - 1，整理后得到(x+3)^2 + (y-1)^2 = 17，说明圆心是(-3, 1)，半径是根号17。混合练习第19题正是这种结构：题目给出x^2 + 6x + y^2 - 2y = 7，然后问切线的方程。如果不会配方，后面所有小题都无法进行。配方的关键是先看x项系数的一半，再加平方，记得减掉多出来的常数，最后移到等号右边就是r^2。

When a question presents an equation such as x^2 + y^2 + 6x - 2y = 7, completing the square must always come first. Group the x terms to form (x+3)^2 - 9 and the y terms to form (y-1)^2 - 1, then rearrange to get (x+3)^2 + (y-1)^2 = 17, which shows the centre is (-3, 1) and the radius is the square root of 17. Question 19 of the mixed exercise uses exactly this structure, asking for tangent equations after giving x^2 + 6x + y^2 - 2y = 7. Without completing the square, none of the later parts can be attempted. The key step is taking half of the x coefficient, squaring it, subtracting the extra constant, and moving the remaining constant to the right-hand side to reveal r squared.

四、三点定圆：列方程组求解 | Finding a Circle from Three Points

已知圆周上三个点的坐标求圆的方程，是本章另一类核心题型。第24题给出A(-4, 0)、B(4, 8)、C(6, 0)三点在圆上，求圆的方程。标准做法是设圆的一般式x^2 + y^2 + 2gx + 2fy + c = 0，把三个点分别代入，得到三个关于g、f、c的方程，联立求解。由于A、C两点都在x轴上，代入后能迅速消元，计算量并不大。另一种方法是作任意两条弦的垂直平分线，交点就是圆心，再用两点距离公式求半径。两种方法都要求细心计算，考试中建议优先用代点法，因为思路固定、不易漏解。

Finding the equation of a circle from three points on its circumference is another core question type in this chapter. Question 24 gives A(-4, 0), B(4, 8) and C(6, 0) and asks for the circle equation. The standard method is to let the general form x^2 + y^2 + 2gx + 2fy + c = 0 stand, substitute each point in turn, and solve the resulting system of three simultaneous equations for g, f and c. Because A and C both lie on the x-axis, substitution eliminates terms quickly and the arithmetic stays light. Alternatively, draw perpendicular bisectors of any two chords; their intersection is the centre, and the radius follows from the distance formula. Both methods reward careful calculation, and in the exam the substitution method is usually safer because the procedure is fixed and less prone to missed cases.

五、中点公式与垂直平分线 | Midpoint Formula and Perpendicular Bisectors

垂直平分线是本章最常考的工具。线段AB的垂直平分线垂直于AB，并且经过AB的中点，如果AB的斜率为m，则垂直平分线的斜率为-1/m。第25题给出A(-7, 7)、B(1, 9)、C(3, 1)、D(-7, 1)四点共圆，要求先求AB和CD的垂直平分线方程，再求圆的方程。解题时先算中点坐标，用两点间斜率公式算出原线段斜率，取负倒数得到垂直平分线斜率，再用点斜式写出方程。两条垂直平分线的交点就是圆心，这是证明四点共圆和确定圆心位置的通用套路。

The perpendicular bisector is the most frequently tested tool in this chapter. The perpendicular bisector of segment AB is perpendicular to AB and passes through its midpoint; if the gradient of AB is m, then the gradient of the bisector is -1/m. Question 25 states that A(-7, 7), B(1, 9), C(3, 1) and D(-7, 1) lie on one circle and asks you to find the perpendicular bisectors of AB and CD first, then the circle equation. The procedure is to calculate the midpoint, find the gradient of the original segment with the two-point formula, take the negative reciprocal for the bisector gradient, and write the line with the point-slope form. The intersection of two perpendicular bisectors is the centre, which is the universal strategy for proving that four points are concyclic and for locating the centre.

六、切线问题：半径与切线垂直 | Tangent Problems: Radius Perpendicular to Tangent

过圆上一点作切线，切线与过该点的半径互相垂直，所以两条直线的斜率乘积等于-1。第14题给出点A(-3, -7)在圆心为(5, 1)的圆上，求过A的切线方程。先求半径CA的斜率，取负倒数得到切线斜率，再用点斜式写出切线方程。混合练习第18题更进一步，已知两条切线的斜率，反求切点坐标：先设切点为(x, y)，利用圆心到切点连线斜率与切线斜率垂直，结合点在圆上，联立方程解出切点，再写切线方程。这类题在Edexcel真题中经常以8分大题出现，步骤分给得很足。

A tangent at a point on a circle is perpendicular to the radius drawn to that point, so the product of their gradients is -1. Question 14 gives point A(-3, -7) on a circle with centre (5, 1) and asks for the tangent at A. First find the gradient of radius CA, take the negative reciprocal for the tangent gradient, then use the point-slope form. Question 18 of the mixed exercise goes further: given the gradients of two tangents, work back to the points of contact. Set the contact point as (x, y), use the fact that the radius and tangent gradients are perpendicular, combine this with the point lying on the circle, and solve the simultaneous equations to locate the contact point before writing the tangent equations. Questions of this type regularly appear as 8-mark problems in Edexcel papers, with generous method marks.

七、弦的性质与圆心到弦的距离 | Chord Properties and Distance from Centre to Chord

圆的弦有一个重要性质：圆心到弦的垂线平分这条弦，反过来，弦的垂直平分线一定经过圆心。第15题给出A(3, 7)、B(5, 1)两点在圆上，AC是直径，要求三角形ABC的面积。解题关键是利用M是AB的中点、CM垂直于AB这两个条件。第17题也用了同样的结构：M是弦AB的中点，直线l经过M和圆心C，所以l就是AB的垂直平分线。遇到弦的问题，先画圆心到弦中点的连线，把它标成直角，整个图形的关系就会清晰很多。这是几何直观层面最重要的一个技巧。

Chords of a circle obey an important property: the perpendicular from the centre to a chord bisects the chord, and conversely the perpendicular bisector of a chord always passes through the centre. Question 15 gives A(3, 7) and B(5, 1) on a circle, states that AC is a diameter, and asks for the area of triangle ABC. The key is to use the facts that M is the midpoint of AB and that CM is perpendicular to AB. Question 17 uses the same structure: M is the midpoint of chord AB, and the line l through M and the centre C is therefore the perpendicular bisector of AB. Whenever a chord appears, draw the line from the centre to the midpoint of the chord and mark the right angle; the whole diagram becomes much easier to read. This is the single most useful visual technique in the chapter.

八、两圆相交与公共弦方程 | Two Intersecting Circles and the Common Chord

两个圆相交时，把两个圆的方程相减，得到的直线方程就是公共弦所在直线，这条直线经过两个交点。混合练习第26题给出两圆(x-5)^2 + (y-3)^2 = 20和(x-10)^2 + (y-8)^2 = 10，圆心分别为A和B，两圆交于P、Q两点，要求公共弦PQ所在直线的方程。做法是先把两个标准式展开成一般式，再相减消去x^2和y^2项，剩下的线性方程ax + by + c = 0就是答案。求出直线后再与其中一个圆联立，就能解出P、Q坐标，最后求四边形APBQ的面积时，注意到它关于AB轴对称，可以用两个三角形面积相加处理。

When two circles intersect, subtracting their equations gives the equation of the common chord, the line that passes through both points of intersection. Question 26 of the mixed exercise gives two circles (x-5)^2 + (y-3)^2 = 20 and (x-10)^2 + (y-8)^2 = 10 with centres A and B, intersecting at P and Q, and asks for the equation of the line PQ. Expand both standard forms into general form, subtract to eliminate the x squared and y squared terms, and the remaining linear equation ax + by + c = 0 is the answer. Solving this line together with one circle yields the coordinates of P and Q. For the final part, the area of quadrilateral APBQ, notice that the shape is symmetric about AB, so the area can be split into two triangles and summed.

九、圆与直线的位置关系 | Position of a Line Relative to a Circle

判断一条直线与圆相交、相切还是相离，方法是把直线方程代入圆的方程，得到一个一元二次方程，然后看判别式：判别式大于0说明有两个交点，等于0说明相切，小于0说明不相交。第20题中，直线x + y - 21 = 0与圆心为(6, 9)、半径为根号50的圆相交于P、Q，就是典型的代入法题目。求出交点后，第20题的后续小题要求分别写出P、Q两点处的切线方程，以及弦PQ的垂直平分线方程。这里垂直平分线一定经过圆心，而且与弦垂直，所以它的斜率与直线PQ的斜率互为负倒数，直接用过圆心和斜率写方程即可。

To decide whether a line intersects, touches, or misses a circle, substitute the line equation into the circle equation to obtain a quadratic, then inspect the discriminant. A positive discriminant means two intersection points, zero means the line is tangent, and a negative value means no intersection. In question 20, the line x + y - 21 = 0 cuts the circle with centre (6, 9) and radius the square root of 50 at points P and Q, a textbook substitution problem. Once the intersections are known, the later parts ask for the tangent equations at P and Q and the perpendicular bisector of chord PQ. That bisector always passes through the centre and is perpendicular to the chord, so its gradient is the negative reciprocal of the gradient of PQ, and the equation follows directly from the centre and that gradient.

十、混合练习典型题拆解 | Worked Examples from the Mixed Exercise

第23题要求证明R(-4, 3)、S(7, 4)、T(8, -7)三点所在的圆以RT为直径。证明直径有三种常用方法：第一种，验证角RST是直角，即两条半径方向上的斜率乘积为-1，由圆周角定理可知RT是直径；第二种，算出RT的中点，再验证S到中点的距离等于半径；第三种，直接证明三角形RST是直角三角形，斜边就是直径。第24题和第25题则分别练习三点定圆和垂直平分线求圆心。建议把这几道题当作本章的自我检测题，限时30分钟完成，对照评分标准给自己打分，能快速暴露薄弱环节。

Question 23 asks you to prove that the circle through R(-4, 3), S(7, 4) and T(8, -7) has RT as its diameter. Three common proofs exist. First, verify that angle RST is a right angle by checking that the product of the gradients of RS and ST is -1; by the angle in a semicircle theorem, RT must then be a diameter. Second, compute the midpoint of RT and check that S is exactly one radius away from it. Third, show directly that triangle RST is right-angled, so the hypotenuse is the diameter. Questions 24 and 25 practise finding circles from three points and using perpendicular bisectors to locate the centre. Treat these problems as a self-test for the chapter: allow thirty minutes, mark yourself against the official mark scheme, and the exercise will quickly expose your weak areas.

十一、常见错误与考试陷阱 | Common Mistakes and Exam Traps

本章丢分集中在四个地方。第一，配方时忘记减去加上的常数，导致半径算错。第二，取斜率的负倒数时只变符号不变大小，或者把垂直平分线直接当成过原点的直线。第三，代入三个点求g、f、c时计算粗心，一个符号错全题错，建议每步都回代检验。第四，题目要求切线方程写成ax + by + c = 0的形式，不少同学写成y = mx + c就被扣分，Edexcel的评分标准对答案形式有明确要求。另外要注意，两个圆的方程相减得到的是公共弦方程，而不是圆心连线的方程，这两者概念上完全不同。

Marks are lost in four typical places in this chapter. First, completing the square without subtracting the added constant, which corrupts the radius. Second, taking the negative reciprocal of a gradient but only changing the sign, or treating a perpendicular bisector as if it passed through the origin. Third, careless arithmetic when substituting three points to solve for g, f and c, where one sign error ruins the whole question; substitute back at every step to check. Fourth, the question often demands the tangent equation in the form ax + by + c = 0, and students who write y = mx + c lose marks because Edexcel mark schemes specify the required form. Also note that subtracting two circle equations gives the common chord, not the line joining the centres, and the two concepts are entirely different.

十二、考前复习计划与刷题建议 | Revision Plan and Practice Strategy

圆的章节建议按三个层次复习。第一层，熟记标准式、一般式、中点公式、斜率乘积为-1这四个核心工具，每天用10分钟默写一遍。第二层，把课本第135页的混合练习全部做一遍，重点做第15、18、19、23、26题，这些题目覆盖了本章所有题型，做完后对照答案批改并整理错题本。第三层，从Edexcel历年真题中收集圆章节的题目，集中训练8分大题，练习把步骤写完整，注意保留根号形式的精确值。圆的题目计算量适中但步骤环环相扣，只要每一步都写清楚，拿到满分并不难。

Revise this chapter in three layers. In layer one, memorise the four core tools: the standard form, the general form, the midpoint formula, and the rule that perpendicular gradients multiply to -1, and rewrite them from memory for ten minutes each day. In layer two, complete the whole mixed exercise on page 135, paying special attention to questions 15, 18, 19, 23 and 26, which together cover every question type in the chapter, then mark your work and build an error log. In layer three, collect circle questions from past Edexcel papers and drill the 8-mark problems, practising full written steps and leaving answers as exact surds where required. The calculations are moderate and the steps chain tightly together; as long as every step is written clearly, full marks are well within reach.

需要一对一A-Level数学辅导？
16621398022
同微信 | 免费咨询课程规划
访问 aleveler.com 了解更多

更多咨询请联系16621398022（同微信）