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

📚 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。

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1. The Problem Statement | 题目陈述

The line \(y = 2x – k\) and the circle \(x^{2} + y^{2} = 5\) 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 \(x^{2} + y^{2} = 5\)。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 \(x^{2} + y^{2} = 5\) has centre \((0,0)\) and radius \(\sqrt{5}\). The line has gradient \(2\) and an unknown intercept \(-k\).

圆 \(x^{2} + y^{2} = 5\) 的圆心为 \((0,0)\)，半径为 \(\sqrt{5}\)。直线的斜率为 \(2\)，截距为未知的 \(-k\)。

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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 \(x^{2} + y^{2} = 5\).

将 \(y = 2x – k\) 代入 \(x^{2} + y^{2} = 5\)。

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。

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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 \(y\)。

\(x^{2} + (2x – k)^{2} = 5\)

Expand the squared bracket carefully.

仔细展开平方括号。

\(x^{2} + 4x^{2} – 4kx + k^{2} = 5\)

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。

\(5x^{2} – 4kx + k^{2} – 5 = 0\)

This is the quadratic that governs the intersection points.

这就是决定交点的二次方程。

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4. Step 2: Extract the Coefficients | 第二步：提取系数

Compare the quadratic with the standard form \(ax^{2} + bx + c = 0\).

将二次方程与标准形式 \(ax^{2} + bx + c = 0\) 进行比较。

Coefficient	Value
\(a\)	\(5\)
\(b\)	\(-4k\)
\(c\)	\(k^{2} – 5\)

Notice that the coefficient \(c\) is \(k^{2} – 5\), not \(k^{2}\). The constant term must include the \(-5\) moved from the right-hand side.

注意，常数项 \(c\) 是 \(k^{2} – 5\)，而不是 \(k^{2}\)。常数项必须包含从等号右边移过来的 \(-5\)。

A common error is to write \(c = k^{2}\), which leads to a completely wrong discriminant.

一个常见错误是把 \(c\) 写成 \(k^{2}\)，这会导致判别式完全错误。

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5. Step 3: Write Down the Discriminant | 第三步：写出判别式

Recall the discriminant formula for the quadratic \(ax^{2} + bx + c = 0\).

回顾二次方程 \(ax^{2} + bx + c = 0\) 的判别式公式。

\(\Delta = b^{2} – 4ac\)

Substitute \(a = 5\), \(b = -4k\), \(c = k^{2} – 5\).

将 \(a = 5\)，\(b = -4k\)，\(c = k^{2} – 5\) 代入。

\(\Delta = (-4k)^{2} – 4 \times 5 \times (k^{2} – 5)\)

Simplify term by term.

逐项化简。

\(\Delta = 16k^{2} – 20(k^{2} – 5)\)

\(\Delta = 16k^{2} – 20k^{2} + 100\)

\(\Delta = -4k^{2} + 100\)

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。

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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。

\(\Delta = 0\)

Set our expression equal to zero.

令我们的表达式等于零。

\(-4k^{2} + 100 = 0\)

Rearrange and solve.

移项并求解。

\(4k^{2} = 100\)

\(k^{2} = 25\)

\(k = \pm 5\)

Therefore, the two possible values are \(k = 5\) and \(k = -5\).

因此，两个可能的值为 \(k = 5\) 和 \(k = -5\)。

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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 \(\sqrt{5}\) 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 \(\sqrt{5}\) 的圆恰好有两条具有该斜率的切线。

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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。

\(d = \frac{|C|}{\sqrt{A^{2} + B^{2}}} = \frac{| – k|}{\sqrt{2^{2} + (-1)^{2}}} = \frac{|k|}{\sqrt{5}}\)

For tangency, set this distance equal to the radius \(\sqrt{5}\).

相切时，令这个距离等于半径 \(\sqrt{5}\)。

\(\frac{|k|}{\sqrt{5}} = \sqrt{5}\)

\(|k| = 5\)

This confirms \(k = \pm 5\), matching the discriminant method perfectly.

这证实了 \(k = \pm 5\)，与判别式法的结果完全一致。

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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
\(\Delta > 0\)	Two distinct roots	Two points: the line cuts the circle
\(\Delta = 0\)	One repeated root	One point: the line is tangent
\(\Delta < 0\)	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。

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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 \(ax^{2} + bx + c = 0\) before reading off coefficients.

• Computing \((-4k)^{2}\) incorrectly as \(-16k^{2}\). The square of a negative number is positive.

• Using \(\Delta > 0\) instead of \(\Delta = 0\). Read the question: ‘exactly one point’ means equality.

• Stopping at \(k^{2} = 25\) 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 \(ax^{2} + bx + c = 0\) 的形式。

• 把 \((-4k)^{2}\) 错误地算成 \(-16k^{2}\)。负数的平方是正数。

• 使用 \(\Delta > 0\) 而不是 \(\Delta = 0\)。仔细读题：“恰好一个点”意味着相等。

• 解到 \(k^{2} = 25\) 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。

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11. Exam-Style Variant | 考试变式题

Try this similar question: the line \(y = 2x + c\) is tangent to the circle \(x^{2} + y^{2} = 4\). Find the possible values of \(c\).

试试这道类似的题目：直线 \(y = 2x + c\) 与圆 \(x^{2} + y^{2} = 4\) 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。

\(x^{2} + (2x + c)^{2} = 4\)

\(x^{2} + 4x^{2} + 4cx + c^{2} = 4\)

\(5x^{2} + 4cx + c^{2} – 4 = 0\)

Set the discriminant to zero.

令判别式等于零。

\((4c)^{2} – 4 \times 5 \times (c^{2} – 4) = 0\)

\(16c^{2} – 20c^{2} + 80 = 0\)

\(-4c^{2} + 80 = 0\)

\(c^{2} = 20\)

\(c = \pm 2\sqrt{5}\)

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。

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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 \(x^{2} + y^{2} = 2\).

• Question 2: The line \(y = 3x + c\) intersects the circle \(x^{2} + y^{2} = 10\) at exactly one point. Find \(c\).

• Question 3: Given that the line \(y = mx + 4\) is tangent to the circle \(x^{2} + y^{2} = 4\), find the possible values of \(m\).

• 题目1：求 \(k\) 的值，使直线 \(y = x + k\) 与圆 \(x^{2} + y^{2} = 2\) 相切。

• 题目2：直线 \(y = 3x + c\) 与圆 \(x^{2} + y^{2} = 10\) 恰好交于一个点，求 \(c\)。

• 题目3：已知直线 \(y = mx + 4\) 与圆 \(x^{2} + y^{2} = 4\) 相切，求 \(m\) 的可能值。

Answers: Question 1: \(k = \pm 2\). Question 2: \(c = \pm 10\). Wait, check this carefully: substituting gives \(10x^{2} + 6cx + c^{2} – 10 = 0\), so \(\Delta = 36c^{2} – 40(c^{2} – 10) = -4c^{2} + 400 = 0\), hence \(c = \pm 10\). Question 3: \(m = \pm \sqrt{3}\): start from \(x^{2} + (mx + 4)^{2} = 4\), which simplifies to \((1 + m^{2})x^{2} + 8mx + 12 = 0\), so \(\Delta = 64m^{2} – 48(1 + m^{2}) = 16m^{2} – 48 = 0\), giving \(m^{2} = 3\).

答案：题目1：\(k = \pm 2\)。题目2：\(c = \pm 10\)。请仔细检验：代入得 \(10x^{2} + 6cx + c^{2} – 10 = 0\)，因此 \(\Delta = 36c^{2} – 40(c^{2} – 10) = -4c^{2} + 400 = 0\)，所以 \(c = \pm 10\)。题目3：\(m = \pm \sqrt{3}\)：从 \(x^{2} + (mx + 4)^{2} = 4\) 出发，化简为 \((1 + m^{2})x^{2} + 8mx + 12 = 0\)，所以 \(\Delta = 64m^{2} – 48(1 + m^{2}) = 16m^{2} – 48 = 0\)，得到 \(m^{2} = 3\)。

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