Intersections of Lines and Circles: Position Relationship Analysis | 直线与圆的交点及位置关系判断

📚 Intersections of Lines and Circles: Position Relationship Analysis | 直线与圆的交点及位置关系判断

In coordinate geometry, the positional relationship between a straight line and a circle is a fundamental topic. It tests not only algebraic manipulation but also geometric reasoning. We will explore how to determine whether a line intersects, touches, or misses a circle, and how to find the exact intersection points.

在平面解析几何中，直线与圆的位置关系是基础核心考点。它不仅考查代数运算能力，更检验几何直观思维。我们将系统学习如何判断一条直线与圆相交、相切还是相离，并掌握求解交点坐标的方法。

1. Standard Equations of Lines and Circles | 直线与圆的标准方程

A straight line can be written in the form y = mx + c (slope-intercept form) or ax + by + c = 0 (general form). A circle with centre (a, b) and radius r has the equation (x – a)² + (y – b)² = r².

直线的方程可表示为 y = mx + c（斜截式）或 ax + by + c = 0（一般式）。圆心为 (a, b)、半径为 r 的圆的标准方程为 (x – a)² + (y – b)² = r²。

Circle: (x – a)² + (y – b)² = r²

When a circle is centred at the origin, the equation simplifies to x² + y² = r².

当圆心位于原点时，方程简化为 x² + y² = r²。

2. Three Possible Position Relationships | 三种位置关系

A line and a circle can have exactly three distinct positional relationships. If the distance from the circle’s centre to the line is greater than the radius, the line lies entirely outside the circle, and there are no intersection points. If the distance equals the radius, the line touches the circle at exactly one point, called the tangent point. If the distance is less than the radius, the line cuts through the circle at two distinct points.

直线与圆的位置关系仅有三种。若圆心到直线的距离大于半径，则直线在圆外，没有交点；若距离等于半径，则直线与圆相切，只有一个切点；若距离小于半径，则直线与圆相交，有两个不同交点。

Distance > r: No intersection, the line is external (相离)
Distance = r: One intersection, the line is tangent (相切)
Distance < r: Two intersections, the line is secant (相交)

3. Algebraic Approach: Substitution and Discriminant | 代数方法：代入法与判别式

The most direct algebraic method is to substitute the line equation into the circle equation, producing a quadratic equation in one variable. The discriminant Δ = b² – 4ac then determines the number of intersection points.

最直接的代数方法是将直线方程代入圆方程，得到关于单个变量的一元二次方程。判别式 Δ = b² – 4ac 的值决定了交点的个数。

Δ > 0 → two intersections ｜ Δ = 0 → one intersection ｜ Δ < 0 → no intersection

Suppose the line is y = mx + c and the circle is (x – a)² + (y – b)² = r². Replacing y with mx + c yields a quadratic in x. The discriminant of this quadratic tells us the number of intersection points.

设直线方程为 y = mx + c，圆方程为 (x – a)² + (y – b)² = r²。将 y 用 mx + c 代替，得到关于 x 的一元二次方程。计算其判别式即可判断交点个数。

4. Worked Example: Using the Discriminant | 判别法实例

Example: Determine whether the line y = 2x + 1 intersects the circle x² + y² = 5.

例：判断直线 y = 2x + 1 与圆 x² + y² = 5 的位置关系。

Substitute y = 2x + 1 into x² + y² = 5:

将 y = 2x + 1 代入 x² + y² = 5：

x² + (2x + 1)² = 5 → x² + 4x² + 4x + 1 = 5 → 5x² + 4x – 4 = 0

The discriminant is Δ = 4² – 4 × 5 × (-4) = 16 + 80 = 96 > 0. Since Δ > 0, the line intersects the circle at two distinct points.

计算判别式：Δ = 4² – 4 × 5 × (-4) = 16 + 80 = 96。由于 Δ > 0，直线与圆相交于两个不同点。

5. Geometric Approach: Distance from Centre to Line | 几何方法：圆心到直线的距离

For a line written in general form ax + by + c = 0, the perpendicular distance from a point (x₁, y₁) to the line is given by the formula:

对于一般式直线 ax + by + c = 0，点 (x₁, y₁) 到直线的垂线距离公式为：

d = |ax₁ + by₁ + c| / √(a² + b²)

In the context of a circle, we take the centre (a, b) as the point. The distance d obtained is then compared with the radius r.

在圆中，我们将圆心 (a, b) 视为该点。求得的距离 d 再与半径 r 比较。

d > r: no intersection, line is outside the circle
d = r: line is tangent to the circle
d < r: line crosses the circle at two points

6. Worked Example: Using the Distance Formula | 距离公式实例

Example: Determine the position of the line 3x – 4y + 10 = 0 relative to the circle with centre (0, 0) and radius 2.

例：判断直线 3x – 4y + 10 = 0 与圆心 (0, 0)、半径 2 的圆的位置关系。

Here a = 3, b = -4, c = 10, and the centre is (0, 0). Apply the distance formula:

这里 a = 3，b = -4，c = 10，圆心为 (0, 0)。代入距离公式：

d = |3×0 + (-4)×0 + 10| / √(3² + (-4)²) = |10| / √25 = 10 / 5 = 2

Since d = 2 = r, the line is tangent to the circle. It touches at exactly one point.

由于 d = 2 = r，直线与圆相切，仅有一个公共点。

7. Finding Intersection Points: General Procedure | 求交点坐标的一般步骤

To find the coordinates of intersection points, solve the line and circle equations simultaneously. Use substitution to create a quadratic equation, then solve for the x-values. Substitute back into the line equation to obtain the corresponding y-values.

求交点坐标需联立直线与圆的方程。用代入法得到一元二次方程，求解 x 值，再代回直线方程求出对应的 y 值。

Step 1: Express y in terms of x from the line equation (if the line is vertical, express x as a constant).

步骤一：由直线方程将 y 表示为 x 的表达式（若直线垂直，则将 x 表示为常数）。

Step 2: Substitute into the circle equation and simplify to a quadratic a x² + b x + c = 0.

步骤二：代入圆方程并化简为 a x² + b x + c = 0 的形式。

Step 3: Solve the quadratic to find x-coordinates; then find y-coordinates.

步骤三：解二次方程得 x 坐标，再求对应的 y 坐标。

8. Special Case: The Tangent Line | 特殊情况：切线

When the discriminant equals zero, the line is tangent to the circle. The quadratic equation has exactly one repeated root. The single intersection point is the point of tangency. The tangent line is always perpendicular to the radius at that point.

当判别式等于零时，直线与圆相切。此时二次方程有唯一重根，该交点即为切点。切线在切点处始终与半径垂直。

For a circle x² + y² = r² and a point (x₁, y₁) on the circle, the tangent equation is xx₁ + yy₁ = r².

对于圆 x² + y² = r² 上的点 (x₁, y₁)，切线方程为 xx₁ + yy₁ = r²。

Tangent at (x₁, y₁): x·x₁ + y·y₁ = r²

9. Finding the Chord Length | 求弦长

When a line intersects a circle at two points, the segment connecting these two points is called a chord. If we know the perpendicular distance d from the centre to the line, the chord length can be found using the right triangle formed by the radius, the perpendicular, and half the chord.

当直线与圆相交于两点时，两点间的线段称为弦。若已知圆心到直线的垂距 d，可利用半径、垂距和半弦长构成的直角三角形求解弦长。

Chord length = 2√(r² – d²)

This formula is elegant and efficient, avoiding the need to find the actual intersection coordinates.

该公式简洁高效，无需实际求出交点坐标即可得到弦长。

10. Worked Example: Chord Length Calculation | 弦长计算实例

Example: The line y = x + 3 cuts the circle (x – 1)² + (y – 2)² = 10. Find the length of the chord.

例：直线 y = x + 3 与圆 (x – 1)² + (y – 2)² = 10 相交，求弦长。

First, write the line in general form: x – y + 3 = 0. The centre of the circle is (1, 2), and the radius is √10.

首先将直线化为一般式：x – y + 3 = 0。圆心为 (1, 2)，半径 r = √10。

d = |1 – 2 + 3| / √(1² + (-1)²) = |2| / √2 = √2

Now apply the chord length formula:

代入弦长公式：

Chord length = 2√(10 – 2) = 2√8 = 4√2

11. Common Pitfalls and Tips | 常见误区与技巧

Students often confuse the two criteria: the discriminant method and the distance method. Remember that the discriminant comes from the quadratic obtained after substitution, while the distance method requires the line in general form. Also, when using the discriminant, careful expansion is essential; a simple sign error leads to the wrong conclusion.

学生常混淆判别式法与距离法。请记住，判别式来自代入后的二次方程，而距离法要求直线写成一般式。另外，使用判别式时必须仔细展开，一个符号错误就会导致结论错误。

Always check whether the circle is centred at the origin or at another point.
If the line is vertical (x = k), substitution is still straightforward but there is no slope.
When Δ = 0, do not forget that the repeated root gives the x-coordinate of the tangency point.
Use approximate values to verify your conclusion when possible.

注意： 始终确认圆心是否在原点。若直线垂直（x = k），代入法依然直接，但斜率不存在。当 Δ = 0 时，重根给出切点的 x 坐标。可能时用近似值检验结论。

12. Summary and Exam Strategy | 总结与应试策略

The relationship between a line and a circle is determined by comparing the distance from the centre to the line with the radius, or equivalently by examining the discriminant of the quadratic obtained after substitution. Both methods are reliable; choose the one that suits the given equation forms.

直线与圆的位置关系可通过比较圆心到直线的距离与半径来判断，也可以通过代入后的二次方程判别式来判断。两种方法都可靠，可根据题目给出的方程形式灵活选择。

Distance method: d = |ax₀ + by₀ + c| / √(a² + b²)

For tangent problems, remember the perpendicularity between the tangent and the radius. For chord problems, use the half-chord right triangle. Practise both algebraic and geometric approaches to become fluent in solving this class of problems.

处理切线问题时，牢记切线与半径垂直；处理弦长问题时，利用半弦直角三角形。通过练习代数法与几何法，熟悉这类问题的解题套路。

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Intersections of Lines and Circles: Position Relationship Analysis | 直线与圆的交点及位置关系判断