📚 Example 5.6.2: Tangent to a Circle | 圆的切线例题 5.6.2
type of question regularly appears in both AS and A-Level examinations, and mastering it builds essential skills in coordinate geometry and differentiation.
本文详细讲解 AQA A-Level 数学中的经典例题：求圆在某一点的切线。例题 5.6.2 要求我们求圆 x² + y² = 25 在点 (3, 4) 处的切线方程。这类题目在 AS 和 A-Level 考试中经常出现，掌握它有助于培养坐标几何与微分的关键技能。
1. Understanding the Problem | 理解题目
ion of the straight line that just touches the circle at this point without crossing it. In coordinate geometry, such a line is called a tangent.
圆 x² + y² = 25 的圆心在原点 (0, 0)，半径为 5。点 (3, 4) 在圆上，因为 3² + 4² = 9 + 16 = 25。我们需要找到一条直线，它在这一点处恰好与圆相触而不穿过圆。在坐标几何中，这样的直线称为切线。
It is always a good habit to verify that the given point actually lies on the circle. If the point were not on the circle, the question would be inconsistent. This quick check also helps you avoid sign errors later in the calculation.
养成验证给定点确实在圆上的好习惯非常重要。如果点不在圆上，题目本身就不成立。这种快速检验还能帮助你在后续计算中避免符号错误。
2. Key Facts About Tangents to Circles | 关于圆的切线的关键事实
A tangent to a circle is a straight line that touches the circle at exactly one point. At the point of contact, the tangent is perpendicular to the radius drawn from the centre to that point. This perpendicular relationship is the foundation of the geometric method for finding tangent equations.
圆的切线是一条与圆恰好只有一个交点的直线。在切点处，切线与从圆心到该点的半径垂直。这一垂直关系是用几何方法求切线方程的基础。
In Year 2 of AQA A-Level Mathematics, you also learn implicit differentiation, which gives another route to the same result. The two methods reinforce each other: the geometric method is quick for circles centred at the origin, while implicit differentiation works for a much wider range of curves.
在 AQA A-Level 数学第二年，你还会学习隐函数求导，它为同一问题提供了另一种解法。两种方法互为补充：几何方法对于圆心在原点的圆更快捷，而隐函数求导适用于更广泛的曲线类型。
3. Method 1: Using the Radius Perpendicular to the Tangent | 方法一：利用半径垂直于切线
The geometric approach uses two basic steps. First, find the gradient of the radius from the centre of the circle to the given point. Second, use the fact that the tangent gradient is the negative reciprocal of the radius gradient, because the two lines are perpendicular.
几何方法分为两个基本步骤。首先，求出从圆心到给定点的半径斜率。然后，利用切线斜率是半径斜率的负倒数这一事实，因为这两条直线互相垂直。
This method is highly efficient for circles with a simple centre such as (0, 0). It also gives a clear visual understanding of why the tangent equation takes a particular form.
对于圆心为 (0, 0) 这样的简单情况，这种方法非常高效。它还能让你从直观上理解为什么切线方程具有特定的形式。
4. Step 1 - Find the Gradient of the Radius | 步骤一：求半径的斜率
0, 0). The given point is (3, 4). The gradient of the radius is the change in y divided by the change in x. Using the standard gradient formula:
圆 x² + y² = 25 的圆心是 (0, 0)。给定点是 (3, 4)。半径的斜率等于 y 的变化量除以 x 的变化量。使用标准斜率公式：
gradient of radius = (4 - 0) / (3 - 0) = 4/3
The radius therefore rises by 4 units for every 3 units it moves to the right. This positive gradient makes sense because the point (3, 4) is in the first quadrant.
因此，半径每向右移动 3 个单位，就上升 4 个单位。这个正斜率是合理的，因为点 (3, 4) 位于第一象限。
5. Step 2 - Use the Perpendicular Gradient Rule | 步骤二：利用垂直斜率规则
If two straight lines are perpendicular, the product of their gradients is -1. Let mₜ be the gradient of the tangent. Since the radius gradient is 4/3, we have:
如果两条直线互相垂直，那么它们斜率的乘积为 -1。设 mₜ 为切线的斜率。由于半径斜率为 4/3，我们有：
mₜ × (4/3) = -1
Solving for mₜ gives:
解出 mₜ 得到：
mₜ = -3/4
A common mistake here is to forget the negative sign or to take the reciprocal without making it negative. Always remember that perpendicular gradients have opposite signs and are reciprocals of each other.
这里一个常见的错误是忘记负号，或者取了倒数但没有变为负数。务必记住，两条垂直直线的斜率符号相反且互为倒数。
6. Step 3 - Write the Tangent Equation | 步骤三：写出切线方程
m of a straight line, y - y₁ = m(x - x₁), where (x₁, y₁) = (3, 4) and m = -3/4, we obtain:
切线经过点 (3, 4)，斜率为 -3/4。使用直线的点斜式方程 y - y₁ = m(x - x₁)，其中 (x₁, y₁) = (3, 4)，m = -3/4，我们得到：
y - 4 = (-3/4)(x - 3)
To express this in a neater form without fractions, multiply both sides by 4:
为了用一个不含分数的更简洁形式表示，将等式两边乘以 4：
4y - 16 = -3x + 9
Rearranging all terms to one side gives the final tangent equation:
将所有项移到一边，得到最终的切线方程：
3x + 4y = 25
= 25, which is correct. This confirms that the tangent passes through the required point.
我们可以将点 (3, 4) 代入方程来检验答案：3(3) + 4(4) = 9 + 16 = 25，结果正确。这证实了切线经过要求的点。
7. Method 2: Implicit Differentiation | 方法二：隐函数求导
differentiate both sides of the equation with respect to x, treating y as a function of x.
在 AQA A-Level 数学第二年，隐函数求导作为一种强大的技巧被引入，用于在 y 没有显式写成 x 的函数时求曲线的斜率。对于 x² + y² = 25 这样的圆，我们可以对等式两边关于 x 求导，并将 y 视为 x 的函数。
This approach is especially useful
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