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

📚 Intersections of Straight Lines and Circles | 直线与圆的交点

In A-Level Mathematics, one of the core coordinate geometry topics is finding the intersection points between a straight line and a circle. This combines algebraic techniques, especially solving quadratic equations, with geometric insight about tangents, chords, and the discriminant. Understanding how many times a line meets a circle, and under what conditions it becomes a tangent, is essential for exam success and forms the foundation for further study in calculus and analytical geometry.

在A-Level数学中，坐标几何的核心主题之一就是求直线与圆的交点。这结合了代数技巧（尤其是解二次方程）以及关于切线、弦和判别式的几何直观。理解一条直线与圆相交的次数，以及在什么条件下它成为切线，对于考试成功至关重要，并且为微积分和解析几何的进一步学习奠定基础。

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1. Review of Circle and Line Equations | 圆与直线方程回顾

A circle with centre (a, b) and radius r is usually given in standard form: (x – a)² + (y – b)² = r². Sometimes it appears in the general expanded form x² + y² + Dx + Ey + F = 0. A straight line can be written in slope-intercept form y = mx + c or in general linear form ax + by + c = 0. Being comfortable with converting between these forms is the first step toward solving intersection problems efficiently.

圆心为 (a, b)、半径为 r 的圆通常用标准形式 (x – a)² + (y – b)² = r² 给出。有时它会以一般展开式 x² + y² + Dx + Ey + F = 0 出现。直线可以写成斜截式 y = mx + c 或一般线性形式 ax + by + c = 0。熟练地在这些形式之间转换，是高效解决交点问题的第一步。

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2. Solving Simultaneous Equations | 联立方程求解

To find where a line meets a circle, we solve their equations simultaneously. The most common approach is substitution: replace y in the circle’s equation with the expression mx + c (or solve for x if the line is vertical). This yields a quadratic equation in x. The solutions give the x-coordinates of intersection points; substitute back to find the corresponding y-coordinates.

为了找到直线与圆的交点，我们联立求解二者的方程。最常用的方法是代入法：将圆的方程中的 y 替换为表达式 mx + c（如果直线是竖直的，则解出 x）。这样会得到一个关于 x 的二次方程。其解就是交点的 x 坐标；代回即可求出对应的 y 坐标。

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3. The Discriminant and the Number of Intersections | 判别式与交点的个数

Once the substitution has produced a quadratic of the form Ax² + Bx + C = 0, the discriminant Δ = B² – 4AC determines how many real intersection points exist. The geometric interpretation is straightforward: if Δ > 0, the line cuts the circle at two distinct points (a secant line); if Δ = 0, the line touches the circle at exactly one point (a tangent); if Δ < 0, the line does not meet the circle at all.

一旦代入产生一个形如 Ax² + Bx + C = 0 的二次方程，判别式 Δ = B² – 4AC 就决定了存在多少个实交点。其几何解释很直接：如果 Δ > 0，直线与圆相交于两个不同的点（割线）；如果 Δ = 0，直线恰好与圆相切于一点（切线）；如果 Δ < 0，直线与圆不相交。

Δ = B² – 4AC

Discriminant	Number of Intersections	Geometric Meaning
Δ > 0	2 distinct points	Secant line
Δ = 0	1 point (repeated root)	Tangent line
Δ < 0	0 points	Line does not meet circle

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4. Condition for Tangency | 相切的条件

A tangent touches the circle at exactly one point. Algebraically, this means the quadratic equation arising from substitution must have a repeated root, so B² – 4AC = 0. This condition allows us to find unknown parameters for a line (such as its gradient m or y-intercept c) so that it is tangent to a given circle. It is frequently examined in Edexcel papers.

切线与圆恰好相切于一点。在代数上，这意味着由代入得到的二次方程必须有重根，因此 B² – 4AC = 0。这个条件使我们能够求出直线的未知参数（例如斜率 m 或 y 轴截距 c），使其成为给定圆的切线。这一考点在 Edexcel 试卷中经常出现。

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5. Finding the Tangent Equation Given the Point of Contact | 已知切点求切线方程

If we know that a point P(x₁, y₁) lies on both the circle and the tangent, there is a more geometric method: the radius from the centre C(a, b) to P is perpendicular to the tangent. The gradient of CP is (y₁ – b)/(x₁ – a), so the gradient of the tangent is the negative reciprocal, provided x₁ ≠ a. The equation of the tangent can then be written using point-slope form: y – y₁ = m_tangent (x – x₁). This avoids heavy algebra and is often quicker.

如果我们已知点 P(x₁, y₁) 既在圆上又在切线上，那么有一种更几何化的方法：从圆心 C(a, b) 到 P 的半径与切线垂直。CP 的斜率是 (y₁ – b)/(x₁ – a)，因此切线的斜率是其负倒数（只要 x₁ ≠ a）。然后可以利用点斜式写出切线方程：y – y₁ = m_tangent (x – x₁)。这避免了繁重的代数运算，通常更快。

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6. Finding Tangents from an External Point | 从圆外一点引切线

When a point T(x₀, y₀) lies outside the circle, two tangents can be drawn to the circle. The standard method is to let the line through T have equation y – y₀ = m(x – x₀). Substitute this into the circle’s equation and impose the tangency condition Δ = 0. This gives a quadratic in m, yielding two slopes (unless one tangent is vertical, which must be checked separately). The final tangent equations can then be stated.

当点 T(x₀, y₀) 位于圆外时，可以作两条切线到圆上。标准方法是设通过 T 的直线方程为 y – y₀ = m(x – x₀)。将此代入圆的方程，并施加相切条件 Δ = 0。这样就得到一个关于 m 的二次方程，得出两个斜率（除非其中一条切线是竖直的，必须单独检查）。然后就可以写出最终的两条切线方程。

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7. Chord Length and the Midpoint of a Chord | 弦长与弦的中点

When a line intersects a circle at two points A and B, the segment AB is called a chord. We can find its length using the quadratic roots from the substitution method. If x₁ and x₂ are the x-coordinates of A and B, then the chord length L is given by:

L = √(1 + m²) × |x₁ – x₂|

where |x₁ – x₂| = √((x₁ + x₂)² – 4x₁x₂). Vieta’s formulas (sum and product of roots) make this very efficient. The midpoint of the chord has x-coordinate (x₁ + x₂)/2 and corresponding y found from the line equation.

当一条直线与圆交于两点 A 和 B 时，线段 AB 叫做一条弦。我们可以利用代入法得到的二次方程的根来求出弦长。如果 x₁ 和 x₂ 是 A、B 的 x 坐标，那么弦长 L 由下式给出：

L = √(1 + m²) × |x₁ – x₂|

其中 |x₁ – x₂| = √((x₁ + x₂)² – 4x₁x₂)。韦达定理（根的和与积）使计算非常高效。弦的中点的 x 坐标是 (x₁ + x₂)/2，对应的 y 坐标由直线方程求得。

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8. Perpendicular Distance from Centre to Chord | 从圆心到弦的垂直距离

Another powerful technique uses the perpendicular distance from the centre of the circle to the line. Let d be the distance from centre C(a, b) to the line ax + by + c = 0. Then the half-chord length satisfies (L/2)² + d² = r². This avoids solving the quadratic entirely and is especially useful when the line equation is given in general form or when the radius and distance are known.

另一种强大的技巧是利用从圆心到直线的垂直距离。设 d 为圆心 C(a, b) 到直线 ax + by + c = 0 的距离。那么半弦长满足 (L/2)² + d² = r²。这就完全避免了求解二次方程，当直线以一般式给出或者半径和距离已知时特别有用。

d = |a·a_centre + b·b_centre + c| / √(a² + b²)

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9. Geometry of Intersections and the Perpendicular Bisector | 交点的几何与垂直平分线

Geometric properties often simplify problems. The perpendicular from the centre of a circle to a chord bisects the chord. Therefore, the line joining the centre to the midpoint of a chord is perpendicular to the chord. This fact can be used to find the equation of a chord given its midpoint, or to verify that a line is a diameter. Combining algebraic and geometric reasoning is a key skill for higher-grade questions.

几何性质常常能简化问题。从圆心到一条弦的垂线平分该弦。因此，连接圆心和弦的中点的直线垂直于该弦。这一事实可用于已知中点求弦的方程，或验证一条直线是否为直径。将代数与几何推理相结合，是解决较高难度题目的关键技能。

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10. Worked Example 1 – Basic Intersection | 典型例题1 – 基础交点

English Example: Find the points of intersection of the line y = 2x + 1 and the circle x² + y² = 5. Substitute y into the circle: x² + (2x + 1)² = 5 → x² + 4x² + 4x + 1 = 5 → 5x² + 4x – 4 = 0. Solve: discriminant Δ = 4² – 4·5·(-4) = 16 + 80 = 96. x = [-4 ± √96] / (2·5) = [-4 ± 4√6] / 10 = -2/5 ± (2√6)/5. Then y = 2x + 1 gives the corresponding y-values. The two intersection points are thus (-0.4 + 0.4√6, 0.2 + 0.8√6) and (-0.4 – 0.4√6, 0.2 – 0.8√6).

中文例题：求直线 y = 2x + 1 与圆 x² + y² = 5 的交点。将 y 代入圆方程：x² + (2x + 1)² = 5 → x² + 4x² + 4x + 1 = 5 → 5x² + 4x – 4 = 0。求解：判别式 Δ = 4² – 4·5·(-4) = 16 + 80 = 96。x = [-4 ± √96] / (2·5) = [-4 ± 4√6] / 10 = -2/5 ± (2√6)/5。然后 y = 2x + 1 给出对应的 y 值。因此两个交点为 (-0.4 + 0.4√6, 0.2 + 0.8√6) 和 (-0.4 – 0.4√6, 0.2 – 0.8√6)。

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11. Worked Example 2 – Tangent Condition | 典型例题2 – 相切条件

English Example: The line y = mx + 5 is a tangent to the circle x² + y² = 20. Find the possible values of m. Substituting gives x² + (mx + 5)² = 20 → x² + m²x² + 10mx + 25 = 20 → (1 + m²)x² + 10mx + 5 = 0. For tangency, Δ = 0 → (10m)² – 4(1 + m²)(5) = 0 → 100m² – 20 – 20m² = 0 → 80m² – 20 = 0 → m² = 1/4 → m = ±1/2. Both gradients are valid.

中文例题：直线 y = mx + 5 与圆 x² + y² = 20 相切。求 m 的可能取值。代入得到 x² + (mx + 5)² = 20 → x² + m²x² + 10mx + 25 = 20 → (1 + m²)x² + 10mx + 5 = 0。相切要求 Δ = 0 → (10m)² – 4(1 + m²)(5) = 0 → 100m² – 20 – 20m² = 0 → 80m² – 20 = 0 → m² = 1/4 → m = ±1/2。两个斜率都有效。

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12. Common Pitfalls and Exam Tips | 常见错误与考试提示

Students often forget to check the case of a vertical line (equation x = constant) when using y = mx + c as the starting form. Always consider x = k separately if the line could be vertical. Another frequent mistake is misapplying the discriminant – ensure the quadratic is fully simplified before identifying A, B, and C. Also, when computing chord lengths, remember that the formula involves the absolute difference of roots, and the square root factor √(1 + m²) depends on the gradient of the line, not the circle. Finally, if a question asks for the equation of a tangent, state it in the requested form, e.g. ax + by + c = 0 or y = mx + c.

学生经常忘记，当以 y = mx + c 作为出发形式时，需要单独检查竖直直线（方程 x = 常数）的情况。如果直线可能是竖直的，务必将 x = k 单独考虑。另一个常见错误是误用判别式——确保二次方程已经完全简化之后，再确定 A、B、C。此外，在计算弦长时，要记住公式涉及根的绝对差，根号因子 √(1 + m²) 取决于直线的斜率，而与圆无关。最后，如果题目要求写出切线方程，请按要求的形式给出，例如 ax + by + c = 0 或 y = mx + c。

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