📚 Circles and Triangles | 圆与三角形
In Edexcel A-Level Mathematics, questions about circles usually ask you to move between a geometric picture and an algebraic equation, while triangle questions reward fluency with the sine rule, the cosine rule and the standard area formulae. The two topics are deeply connected: every triangle has a unique circumcircle and a unique incircle, and several key formulae link the side lengths, the angles and the radii of those circles.
在爱德思 A-Level 数学中，关于圆的题目通常要求你在几何图形与代数方程之间自由切换；而三角形题目则考查你对正弦定理、余弦定理以及各种面积公式的熟练程度。这两个主题联系紧密：每个三角形都有唯一的外接圆与内切圆，并且有若干核心公式把边长、角度与这两个圆的半径联系起来。
1. Centre-radius form of a circle | 圆的圆心-半径式
A circle is defined as the set of all points in a plane that lie at a fixed distance r from a fixed point C. The fixed point is the centre and the fixed distance is the radius. If C has coordinates (a, b) and P(x, y) is any point on the circle, then the length of CP must equal r.
圆被定义为平面内到定点 C 的距离等于定长 r 的所有点的集合。定点是圆心，定长是半径。如果 C 的坐标为 (a, b)，P(x, y) 是圆上任意一点，那么线段 CP 的长度必须等于 r。
Applying the distance formula and squaring both sides gives the centre-radius form, which is the most examinable version of the circle equation.
应用两点间距离公式并对两边平方，就得到圆心-半径式，这是考试中最常用的圆的方程形式。
(x − a)² + (y − b)² = r²
)² means that b = 5. When the centre is the origin, the equation reduces to x² + y² = r², and when r = 1 the circle is called the unit circle.
从这个形式中可以直接读出圆心为 (a, b)，半径为 r，前提是 r > 0。注意符号：(x + 3)² 表示 a = −3，(y − 5)² 表示 b = 5。当圆心在原点时，方程简化为 x² + y² = r²；当 r = 1 时，该圆称为单位圆。
A useful check list before you answer any circle question:
- Is the equation already in centre-radius form, or must it be rearranged?
- Have you copied the signs of a and b correctly?
- Have you written the radius r rather than r²?
- Does the question ask for the equation of the circle, or for points on it?
2. The general form and its centre and radius | 圆的一般式及其圆心与半径
If you expand the centre-radius form you obtain the general form of a circle, in which the coefficients of x² and y² are equal. This is the form most commonly printed in exam papers, and you must be able to extract the centre and radius from it in one or two lines.
如果把圆心-半径式展开，就得到圆的一般式，其中 x² 与 y² 的系数相等。这是试卷中最常见的呈现形式，你必须能在一两步之内从中读出圆心和半径。
x² + y² + 2gx + 2fy + c = 0
g, −f). The constant term satisfies c = a² + b² − r², which rearranges to give the radius.
把它与 (x − a)² + (y − b)² = r² 比较系数，可得 a = −g、b = −f，因此圆心为 (−g, −f)。常数项满足 c = a² + b² − r²，移项后即可得到半径。
radius = √g² + f² − c
The expression under the square root decides what kind of locus the equation represents. This is a favourite multiple-choice and "explain" question.
根号下的表达式决定了这个方程代表什么类型的轨迹。这是选择题和"说明理由"类问题的常见考点。
| Condition | 条件 | Locus | 轨迹 |
| g² + f² − c > 0 | A genuine circle with radius √g² + f² − c | 真正的圆，半径为 √g² + f² − c |
| g² + f² − c = 0 | A single point (−g, −f) | 一个点 (−g, −f) |
| g² + f² − c < 0 | No real points at all | 不存在任何实数点 |
2y² − 8x + 12y − 6 = 0 becomes x² + y² − 4x + 6y − 3 = 0, giving centre (2, −3) and radius √4 + 9 + 3 = 4.
如果 x² 与 y² 的系数不都是 1，要先用这个公系数除整个方程。例如 2x² + 2y² − 8x + 12y − 6 = 0 可化为 x² + y² − 4x + 6y − 3 = 0，于是圆心为 (2, −3)，半径为 √4 + 9 + 3 = 4。
3. Completing the square | 配方法求圆心和半径
Exam questions frequently give a circle in expanded form with awkward coefficients, so the safest method is completing the square separately for the x terms and the y terms. The method works even when the numbers do not factorise nicely.
考试题目经常给出系数不太漂亮的一般式，因此最稳妥的方法是对 x 项和 y 项分别配方。即使数字不能整除，这个方法依然适用。
x² + y² − 6x + 4y − 12 = 0. Group the terms and move the constant to the right-hand side.
考虑方程 x² + y² − 6x + 4y − 12 = 0。先把同类项分组，并把常数移到等号右边。
(x² − 6x) + (y² + 4y) = 12
)² − 4 = 12, so (x − 3)² + (y + 2)² = 25. Hence the centre is (3, −2) and the radius is 5.
对每个括号配方得到 (x − 3)² − 9 + (y + 2)² − 4 = 12，因此 (x − 3)² + (y + 2)² = 25。所以圆心为 (3, −2)，半径为 5。
istake is to leave the equation as (x − 3)² + (y + 2)² = 12, which gives the wrong radius.
注意括号内被减去的两个常数 −9 和 −4。它们必须被加回等号右边。一个非常常见的错误是把方程留成 (x − 3)² + (y + 2)² = 12，从而得到错误的半径。
A short routine to follow every time:
- Divide by the common coefficient of x² and y² if it is not 1.
- Group x terms and y terms, move the constant to the right.
- Half the coefficient of x, square it, add and subtract it.
- Do the same for y, then collect all constants on the right.
- Read off the centre and take the square root of the right-hand side to get r.
4. Intersections of a circle and a line | 圆与直线的交点
To find where a line meets a circle, substitute the equation of the line into the equation of the circle. This produces a quadratic in one variable, and the discriminant tells you exactly how many intersections there are.
求直线与圆的交点，只需把直线方程代入圆的方程。这会得到一个单变量的二次方程，而判别式正好告诉你交点有几个。
| Discriminant | 判别式 | Meaning | 含义 |
| b² − 4ac > 0 | Two distinct points; the line is a chord | 两个不同交点；直线是一条弦 |
| b² − 4ac = 0 | One repeated root; the line is a tangent | 一个重根；直线与圆相切 |
| b² − 4ac < 0 | No real intersection | 没有实数交点 |
mx + k and a centre C(a, b), the perpendicular distance from the centre to the line is d = |ma − b + k| ÷ √m² + 1 s the same three cases: d < r for a chord, d = r for a tangent, d > r for no intersection.
还有一个更快的几何方法。对于写成 y = mx + k 的直线与圆心 C(a, b)，圆心到直线的垂直距离为 d = |ma − b + k| ÷ √m² + 1。把 d 与 r 相比较，同样得到三种情况：d < r 为相交弦，d = r 为相切，d > r 为无交点。
Two results follow immediately. First, the perpendicular from the centre to a chord bisects that chord. Second, if the chord lies at distance d from the centre, its length is given by the formula below.
由此立刻得到两个结论。第一，从圆心向弦所作的垂线平分这条弦。第二，若弦到圆心的距离为 d，则弦长由下式给出。
chord length = 2√r² − d²
5. Tangents and normals to a circle | 圆的切线与法线
A tangent to a circle touches it at exactly one point, and the key theorem is that the tangent is perpendicular to the radius drawn to the point of contact. The normal at that point is therefore the line through the point and the centre.
圆的切线只与圆相交于一点，核心定理是：切线垂直于过切点的半径。因此该点处的法线就是经过该点与圆心的直线。
For a circle with centre (a, b) and radius r, the tangent at the point (x₁, y₁) on the circle has a particularly neat equation, because the vector from the centre to (x₁, y₁) is perpendicular to the tangent direction.
对于圆心为 (a, b)、半径为 r 的圆，圆上一点 (x₁, y₁) 处的切线方程有一个非常简洁的形式，因为从圆心指向 (x₁, y₁) 的向量垂直于切线方向。
(x₁ − a)(x − a) + (y₁ − b)(y − b) = r²
= 0, the tangent at (x₁, y₁) on the circle is obtained by the standard substitution rule.
在一般式 x² + y² + 2gx + 2fy + c = 0 中，圆上一点 (x₁, y₁) 处的切线可由标准替换规则得到。
xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
To find the tangent from an external point P, use the fact that the tangent length satisfies a right-angled triangle relationship with the radius and the line from P to the centre.
要从圆外一点 P 作切线，可利用切线段、半径以及 P 到圆心的连线所构成的直角三角形关系。
tangent length = √PC² − r²
Typical exam tasks in this area include: showing that a given line is a tangent, finding the equation of the tangent at a stated point, and finding the coordinates of the point of contact when the tangent is given. In every case, start by writing down the centre and the radius.
这一部分的常见考题包括：证明某条直线是切线、求指定点处的切线方程、以及已知切线求切点坐标。无论哪种题型，第一步都是写出圆心和半径。
6. Three points determine a circle | 三点确定一个圆
Any three non-collinear points determine exactly one circle, called the circumcircle of the triangle they form. The centre of that circle is the circumcentre, and it lies at the intersection of the perpendicular bisectors of the sides.
任意三个不共线的点唯一确定一个圆，这个圆就是它们所构成的三角形的外接圆。该圆的圆心是外心，位于三边垂直平分线的交点处。
od substitutes the three points into x² + y² + 2gx + 2fy + c = 0 and solves three linear equations for g, f and c.
有两种可靠的方法。几何法是写出两条边的垂直平分线方程并联立求解。代数法是把三个点代入 x² + y² + 2gx + 2fy + c = 0，得到关于 g、f、c 的三个线性方程并求解。
Worked example. Find the circumcircle of the triangle with vertices A(0, 0), B(4, 0) and C(0, 3). Because the angle at A is 90°, the side BC is a diameter of the circle (Thales' theorem), so the centre is the midpoint of BC, namely (2, 1.5), and the radius is half of BC.
例题：求顶点为 A(0, 0)、B(4, 0)、C(0, 3) 的三角形的外接圆。由于 A 处的角是 90°，所以边 BC 是圆的一条直径（泰勒斯定理），因此圆心是 BC 的中点，即 (2, 1.5)，半径为 BC 的一半。
BC = √4² + 3² = 5, so R = 2.5 and (x − 2)² + (y − 1.5)² = 6.25
0. You can verify the answer instantly by checking that each of the three original points satisfies the final equation.
展开得 x² + y² − 4x − 3y = 0。把原来三个点分别代入最终方程，即可立即检验答案是否正确。
7. The sine rule and the circumradius | 正弦定理与外接圆半径
The sine rule relates each side of a triangle to the sine of the opposite angle, and its extended form reveals a direct link with the circumcircle.
正弦定理把三角形的每条边与其对角的正弦联系起来，而它的推广形式则揭示了与外接圆的直接联系。
a ÷ sin A = b ÷ sin B = c ÷ sin C = 2R
Here R is the radius of the circumcircle of the triangle. This single fact allows you to move from an angle and its opposite side straight to the circumradius, and vice versa, without ever finding the centre coordinates.
这里的 R 就是三角形外接圆的半径。仅凭这一条，就可以从某个角及其对边直接求出外接圆半径，反之亦然，完全不需要求圆心的坐标。
R = a ÷ (2 sin A) = abc ÷ (4Δ)
The symbol Δ stands for the area of the triangle. The second version is especially useful because it links three sides, the area and the circumradius in one relationship. If you know all three sides, you can find the area first and then the circumradius.
符号 Δ 表示三角形的面积。第二个版本特别有用，因为它用一个关系式就把三条边、面积和外接圆半径联系在一起。如果已知三边，可以先求面积，再求外接圆半径。
Typical uses in an exam include finding the radius of the circle passing through three given points, or finding an unknown side when the circumradius and one angle are given. Remember that 2R appears, not R, so divide by 2 at the very end.
考试中的典型用法包括：求过三个已知点的圆的半径，或者在已知外接圆半径和一个角的情况下求未知边。注意公式中是 2R 而不是 R，所以最后一步要除以 2。
8. Area, semi-perimeter and the incircle | 面积、半周长与内切圆
The incircle of a triangle is the circle inside the triangle that is tangent to all three sides. Its centre is the incentre, which is the intersection of the three angle bisectors, and its radius is denoted by r.
三角形的内切圆是位于三角形内部、与三条边都相切的圆。它的圆心是内心，位于三条角平分线的交点处，其半径记为 r。
Introduce the semi-perimeter s, which is half of the perimeter. Then Heron's formula gives the area from the three sides alone.
引入半周长 s，即周长的一半。于是海伦公式仅用三条边就能求出面积。
s = (a + b + c) ÷ 2, Δ = √(s(s − a)(s − b)(s − c))
The incircle produces a beautiful area relationship: the area of the triangle equals the semi-perimeter multiplied by the inradius. This is because the triangle can be split into three smaller triangles, each with height r.
内切圆带来一个漂亮的面积关系：三角形面积等于半周长乘以内切圆半径。这是因为三角形可以分割成三个以 r 为高的小三角形。
Δ = rs, so r = Δ ÷ s = √((s − a)(s − b)(s − c) ÷ s)
CA and AB at points X, Y and Z, then AX = AY = s − a, BX = BZ = s − b and CY = CZ = s − c.
从每个顶点出发的切线长度也值得记住。若内切圆与边 BC、CA、AB 分别相切于 X、Y、Z，则 AX = AY = s − a，BX = BZ = s − b，CY = CZ = s − c。
- The incircle lies inside the triangle and touches all three sides internally.
- The circumcircle passes through all three vertices and may lie partly outside the triangle.
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