📚 Example 6.4.2: The Discriminant and Tangency of a Line and a Circle | 例6.4.2：判别式与直线和圆的相切
In this worked example we investigate the condition under which a straight line and a circle meet at exactly one point. This is one of the most commonly tested applications of the discriminant in AQA A-Level Mathematics.
在这个例题中，我们研究一条直线与一个圆恰好相交于一个点的条件。这是 AQA A-Level 数学中最常考察的判别式应用之一。
1. The Problem Statement | 题目陈述
The line y = 2x - k and the circle x2 + y2 = 5 are given. Find the values of the constant k for which the line and the circle meet at exactly one point.
已知直线 y = 2x - k 和圆 x2 + y2 = 5。求常数 k 的值，使得直线与圆恰好相交于一个点。
We should read this carefully. The word 'exactly' tells us that the line must be tangent to the circle, neither passing through it twice nor missing it entirely.
我们应仔细审题。关键词“恰好”告诉我们，直线必须与圆相切，既不能与圆相交于两个点，也不能完全没有交点。
The circle x2 + y2 = 5 has centre (0,0) and radius √5. The line has gradient 2 and an unknown intercept -k.
圆 x2 + y2 = 5 的圆心为 (0,0)，半径为 √5。直线的斜率为 2，截距为未知的 -k。
2. The Key Idea: Substitution | 关键思路：代入法
To find intersection points of a line and a curve, we substitute the equation of the line into the equation of the circle. This produces a quadratic equation in x.
为了求直线与曲线的交点，我们将直线方程代入圆的方程。这样会得到一个关于 x 的二次方程。
Substitute y = 2x - k into x2 + y2 = 5.
将 y = 2x - k 代入 x2 + y2 = 5。
After substitution, the number of solutions of the quadratic tells us the number of intersection points. A quadratic with two distinct real roots corresponds to two intersection points. One repeated root corresponds to one intersection point, which is tangency. No real roots corresponds to no intersection.
代入后，二次方程解的数量告诉我们交点的数量。有两个不同的实数根对应两个交点；有一个重根对应一个交点，即相切；没有实数根则对应没有交点。
Therefore, our task becomes: find k such that the resulting quadratic has exactly one repeated root.
因此，我们的任务就变成：求 k 的值，使所得的二次方程恰好有一个重根。
3. Step 1: Form the Quadratic Equation | 第一步：构造二次方程
Start with the circle equation and replace y with 2x - k.
从圆的方程出发，用 2x - k 替换 y。
x2 + (2x - k)2 = 5
Expand the squared bracket carefully.
仔细展开平方括号。
x2 + 4x2 - 4kx + k2 = 5
Collect like terms and bring the constant 5 to the left-hand side.
合并同类项，并把常数 5 移到等号左边。
5x2 - 4kx + k2 - 5 = 0
This is the quadratic that governs the intersection points.
这就是决定交点的二次方程。
4. Step 2: Extract the Coefficients | 第二步：提取系数
Compare the quadratic with the standard form ax2 + bx + c = 0.
将二次方程与标准形式 ax2 + bx + c = 0 进行比较。
| Coefficient | Value |
| a | 5 |
| b | -4k |
| c | k2 - 5 |
Notice that the coefficient c is k2 - 5, not k2. The constant term must include the -5 moved from the right-hand side.
注意，常数项 c 是 k2 - 5，而不是 k2。常数项必须包含从等号右边移过来的 -5。
A common error is to write c = k2, which leads to a completely wrong discriminant.
一个常见错误是把 c 写成 k2，这会导致判别式完全错误。
5. Step 3: Write Down the Discriminant | 第三步：写出判别式
Recall the discriminant formula for the quadratic ax2 + bx + c = 0.
回顾二次方程 ax2 + bx + c = 0 的判别式公式。
Δ = b2 - 4ac
Substitute a = 5, b = -4k, c = k2 - 5.
将 a = 5，b = -4k，c = k2 - 5 代入。
Δ = (-4k)2 - 4 × 5 × (k2 - 5)
Simplify term by term.
逐项化简。
Δ = 16k2 - 20(k2 - 5)
Δ = 16k2 - 20k2 + 100
Δ = -4k2 + 100
The discriminant is now a simple expression in k.
判别式现在变成了一个关于 k 的简单表达式。
6. Step 4: Apply the Tangency Condition | 第四步：应用相切条件
For the line to touch the circle at exactly one point, the quadratic must have exactly one repeated root. That happens precisely when the discriminant is zero.
要使直线与圆恰好相切于一个点，二次方程必须有且仅有一个重根。这恰好发生在判别式等于零的时候。
Δ = 0
Set our expression equal to zero.
令我们的表达式等于零。
-4k2 + 100 = 0
Rearrange and solve.
移项并求解。
4k2 = 100
k2 = 25
k = ± 5
Therefore, the two possible values are k = 5 and k = -5.
因此，两个可能的值为 k = 5 和 k = -5。
7. Interpret the Two Lines | 解读两条直线
Substitute the values of k back into the line equation.
将 k 的值代回直线方程。
| Value of k | Line equation |
| k = 5 | y = 2x - 5 |
| k = -5 | y = 2x + 5 |
Both lines have the same gradient 2, so they are parallel to each other. They lie symmetrically on either side of the circle, each touching the circle at a single point.
两条直线的斜率都是 2，因此它们彼此平行。它们对称地位于圆的两侧，各自与圆在唯一一点相切。
This makes sense geometrically: for a fixed gradient, a circle of radius √5 has exactly two tangent lines with that gradient.
这在几何上是合理的：对于固定的斜率，半径为 √5 的圆恰好有两条具有该斜率的切线。
8. Alternative Verification: Perpendicular Distance | 替代验证：垂距法
There is a second method that can be used to check our answer. The distance from the centre of the circle to a tangent line must equal the radius.
还有一种方法可以用来检验我们的答案。从圆心到切线的距离必须等于半径。
Rewrite the line y = 2x - k in the general form Ax + By + C = 0.
将直线 y = 2x - k 改写成一般形式 Ax + By + C = 0。
2x - y - k = 0
The perpendicular distance from the origin to this line is given by the distance formula.
原点到这条直线的垂直距离由距离公式给出。
d = |C|⁄√A2 + B2 = | - k|⁄√22 + (-1)2 = |k|√5
For tangency, set this distance equal to the radius √5.
相切时，令这个距离等于半径 √5。
|k|√5 = √5
|k| = 5
This confirms k = ± 5, matching the discriminant method perfectly.
这证实了 k = ± 5，与判别式法的结果完全一致。
9. Summary of the Discriminant Cases | 判别式各种情况总结
The discriminant of the quadratic formed by substitution always tells us the intersection behaviour.
代入后形成的二次方程的判别式总能告诉我们交点的性质。
| Discriminant | Number of real roots | Intersection with the circle |
| Δ > 0 | Two distinct roots | Two points: the line cuts the circle |
| Δ = 0 | One repeated root | One point: the line is tangent |
| Δ < 0 | No real roots | No points: the line misses the circle |
Memorising this table is useful, but understanding where it comes from is even better. Each root of the quadratic corresponds to an x-coordinate of an intersection point.
记住这张表很有用，但理解它的来源更重要。二次方程的每一个根都对应一个交点的 x 坐标。
10. Common Pitfalls and Exam Advice | 常见错误与应试建议
Students often lose marks on this type of question for a few predictable reasons.
学生在这类题目上失分，通常有几个可以预见的原因。
Forgetting to include the -5 in the constant term c. Always write out the quadratic in the form ax2 + bx + c = 0 before reading off coefficients.
Computing (-4k)2 incorrectly as -16k2. The square of a negative number is positive.
Using Δ > 0 instead of Δ = 0. Read the question: 'exactly one point' means equality.
Stopping at k2 = 25 and writing only k = 5. Remember that a positive and a negative square root both exist.
忘记把 -5 包含在常数项 c 中。在读取系数之前，一定要先把二次方程写成 ax2 + bx + c = 0 的形式。
把 (-4k)2 错误地算成 -16k2。负数的平方是正数。
使用 Δ > 0 而不是 Δ = 0。仔细读题：“恰好一个点”意味着相等。
解到 k2 = 25 就停下来，只写 k = 5。记住正负两个平方根都存在。
In the exam, show every step of the substitution and the simplification of the discriminant. Even if your final values are correct, partial marks are awarded for a clearly structured method.
考试中，要写出代入和化简判别式的每一步。即使最终数值正确，清晰的解题过程也能获得步骤分。
11. Exam-Style Variant | 考试变式题
Try this similar question: the line y = 2x + c is tangent to the circle x2 + y2 = 4. Find the possible values of c.
试试这道类似的题目：直线 y = 2x + c 与圆 x2 + y2 = 4 相切。求 c 的可能值。
Following the same method, substitute into the circle equation.
按照同样的方法，代入圆的方程。
x2 + (2x + c)2 = 4
x2 + 4x2 + 4cx + c2 = 4
5x2 + 4cx + c2 - 4 = 0
Set the discriminant to zero.
令判别式等于零。
(4c)2 - 4 × 5 × (c2 - 4) = 0
16c2 - 20c2 + 80 = 0
-4c2 + 80 = 0
c2 = 20
c = ± 2√5
Again, two tangent lines of gradient 2, each touching the circle of radius 2 at one point.
同样，有两条斜率为 2 的切线，各自与半径为 2 的圆在一点相切。
12. Further Practice | 巩固练习
Work through the following questions on your own, then check your answers.
请独立完成下面的问题，然后核对答案。
Question 1: Find the values of k for which the line y = x + k is tangent to the circle x2 + y2 = 2.
Question 2: The line y = 3x + c intersects the circle x2 + y2 = 10 at exactly one point. Find c.
Question 3: Given that the line y = mx + 4 is tangent to the circle x2 + y2 = 4, find the possible values of m.
题目1：求 k 的值，使直线 y = x + k 与圆 x2 + y2 = 2 相切。
题目2：直线 y = 3x + c 与圆 x2 + y2 = 10 恰好交于一个点，求 c。
题目3：已知直线 y = mx + 4 与圆 x2 + y2 = 4 相切，求 m 的可能值。
Answers: Question 1: k = ± 2. Question 2: c = ± 10. Wait, check this carefully: substituting gives 10x2 + 6cx + c2 - 10 = 0, so Δ = 36c2 - 40(c2 - 10) = -4c2 + 400 = 0, hence c = ± 10. Question 3: m = ± √3: start from x2 + (mx + 4)2 = 4, which simplifies to (1 + m2)x2 + 8mx + 12 = 0, so Δ = 64m2 - 48(1 + m2) = 16m2 - 48 = 0, giving m2 = 3.
答案：题目1：k = ± 2。题目2：c = ± 10。请仔细检验：代入得 10x2 + 6cx + c2 - 10 = 0，因此 Δ = 36c2 - 40(c2 - 10) = -4c2 + 400 = 0，所以 c = ± 10。题目3：m = ± √3：从 x2 + (mx + 4)2 = 4 出发，化简为 (1 + m2)x2 + 8mx + 12 = 0，所以 Δ = 64m2 - 48(1 + m2) = 16m2 - 48 = 0，得到 m2 = 3。
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