📚 Exercise 6I: Circles & Coordinate Geometry | 练习6I：圆与坐标几何
Exercise 6I in the AQA A-Level Mathematics course focuses on coordinate geometry of circles, blending algebraic manipulation with geometric reasoning. This revision guide covers the key techniques needed for the mixed exercise, including finding centre and radius, equations of tangents and chords, and solving intersection problems accurately.
AQA A-Level 数学课程中的练习 6I 主要围绕圆的坐标几何，将代数运算与几何推理相结合。本复习指南涵盖该综合练习所需的关键技巧，包括求圆心与半径、切线与弦的方程以及精确求解相交问题。
1. Standard Form and Completing the Square | 标准形式与配方法
panded form x² + y² + 2gx + 2fy + c = 0, which must be converted by completing the square.
圆的标准方程为 (x - a)² + (y - b)² = r²，其中 (a, b) 为圆心，r 为半径。许多考试题给出的是展开式 x² + y² + 2gx + 2fy + c = 0，必须通过配方法将其转换。
Example: x² + y² - 6x + 8y - 11 = 0
² - 16. The equation becomes (x - 3)² + (y + 4)² = 36, so the centre is (3, -4) and r = 6.
将 x 项和 y 项分组，然后对每个变量配方。对于 x² - 6x，加减 9 得到 (x - 3)² - 9；对于 y² + 8y，加减 16 得到 (y + 4)² - 16。方程变为 (x - 3)² + (y + 4)² = 36，因此圆心为 (3, -4)，半径 r = 6。
- Always keep the equation equal to r² on the right-hand side.
- 配方后等号右侧必须保持为 r² 的形式。
- Check the signs carefully: (x + 4)² means centre x-coordinate is -4.
- 仔细检查符号：(x + 4)² 表示圆心 x 坐标为 -4。
2. Centre and Radius from General Equation | 从一般方程求圆心与半径
For the general form x² + y² + 2gx + 2fy + c = 0, the centre is (-g, -f) and the radius is √g² + f² - c. This formula saves time but requires the equation to have coefficients of x² and y² equal to 1.
对于一般式 x² + y² + 2gx + 2fy + c = 0，圆心为 (-g, -f)，半径为 √g² + f² - c。该公式能节省时间，但要求 x² 和 y² 的系数均为 1。
Centre = (-g, -f), r = √g² + f² - c
0, divide everything by 2 first. Then g = -2, f = 3, c = -3, giving centre (2, -3) and r = √4 + 9 + 3 = √16 = 4.
如果方程为 2x² + 2y² - 8x + 12y - 6 = 0，应先将每一项除以 2。然后 g = -2，f = 3，c = -3，得到圆心 (2, -3) 和 r = √4 + 9 + 3 = √16 = 4。
- Remember: if g² + f² - c is negative, the equation does not represent a real circle.
- 记住：如果 g² + f² - c 为负，则该方程不表示一个真实的圆。
- Always simplify coefficients to 1 before applying the formula.
- 应用公式前必须先将 x² 和 y² 的系数化为 1。
3. Equation of a Circle Given Centre and Radius | 已知圆心与半径求圆的方程
e touches a coordinate axis, the radius equals the absolute distance from the centre to that axis.
要写出圆的方程，只需将圆心 (a, b) 和半径 r 代入 (x - a)² + (y - b)² = r²。如果圆与坐标轴相切，半径等于圆心到该坐标轴的绝对距离。
le, a circle with centre (-2, 5) and passing through the origin has radius √((-2)² + 5²) = √29. Its equation is (x + 2)² + (y - 5)² = 29.
例如，圆心为 (-2, 5) 且过原点的圆，其半径为 √((-2)² + 5²) = √29。它的方程为 (x + 2)² + (y - 5)² = 29。
- If the circle touches the x-axis, r = |b|; if it touches the y-axis, r = |a|.
- 若圆与 x 轴相切，r = |b|；若与 y 轴相切，r = |a|。
- Use Pythagoras to find r from the centre and a point on the circumference.
- 利用勾股定理，从圆心和圆周上一点求 r。
4. Tangents and the Radius Property | 切线与半径垂直性质
A tangent to a circle is perpendicular to the radius at the point of contact. This property is the key to finding tangent equations: first find the gradient of the radius, then use the negative reciprocal for the tangent.
圆的切线在切点处与半径垂直。这一性质是求切线方程的关键：先求出半径的斜率，然后取负倒数得到切线斜率。
m_tangent = -1 / mradius
-4/3. If the radius is horizontal, the tangent is vertical and has equation x = constant.
如果半径的斜率为 3/4，则切线斜率为 -4/3。如果半径是水平的，则切线是竖直的，其方程为 x = 常数。
- Always state that the tangent is perpendicular to the radius, citing this property in your working.
- 解题时必须明确写出“切线与半径垂直”这一性质。
- For vertical tangents, the gradient is undefined, so write the equation as x = a.
- 对于竖直切线，斜率无定义，应写成 x = a 的形式。
5. Finding Tangent Equations | 求切线方程
o the point, then use the negative reciprocal and the point-slope form y - y₁ = m(x - x₁).
求给定点处的切线方程时，先计算圆心到该点的半径斜率，然后取负倒数，并利用点斜式 y - y₁ = m(x - x₁)。
) = 4/3. Tangent gradient = -3/4. Equation: y - 2 = -3/4 (x - 4), giving 3x + 4y - 20 = 0.
例题：圆 (x - 1)² + (y + 2)² = 25，在点 P(4, 2) 处的切线。圆心 C = (1, -2)。CP 的斜率 = (2 - (-2))/(4 - 1) = 4/3。切线斜率 = -3/4。方程为 y - 2 = -3/4 (x - 4)，化为 3x + 4y - 20 = 0。
- Always verify that the point lies on the circle before finding the tangent.
- 求切线前应先验证该点确实在圆上。
- Rearrange the final equation into the form ax + by + c = 0 if required.
- 如需将结果整理为 ax + by + c = 0 的形式。
6. Chord Properties and Perpendicular Bisectors | 弦的性质与垂直平分线
A chord is a line segment joining two points on the circle. The perpendicular bisector of any chord passes through the centre of the circle. This property is especially useful for finding the centre from three points on the circumference.
弦是连接圆上两点的线段。任何弦的垂直平分线都经过圆心。这一性质在由圆周上三点求圆心时特别有用。
To find the centre, take two chords, find their midpoints, and write the equations of the perpendicular bisectors. Their intersection is the centre.
求圆心时，取两条弦，找出它们的中点，并写出垂直平分线的方程。两条垂直平分线的交点即为圆心。
- istance from the centre to a chord can be found using Pythagoras: d² + (half chord)² = r².
- 圆心到弦的距离可用勾股定理计算：d² + (半弦长)² = r²。
- If a chord is horizontal or vertical, its perpendicular bisector is simple to find.
- 如果弦是水平或竖直的，其垂直平分线很容易求出。
7. Intersections of Lines and Circles | 直线与圆的交点
To find the intersection points of a line and a circle, substitute the line equation into the circle equation to form a quadratic in one variable. The discriminant of this quadratic tells you how many intersections exist.
求直线与圆的交点时，将直线方程代入圆的方程，得到关于一个变量的二次方程。该二次方程的判别式可告诉你交点的个数。
Δ = b² - 4ac
at two points. If Δ = 0, the line is tangent. If Δ < 0, the line does not meet the circle.
如果 Δ > 0，直线与圆相交于两点；如果 Δ = 0，直线与圆相切；如果 Δ < 0，直线与圆不相交。
- Substitute y = mx + c into the circle equation and simplify carefully.
- 将 y = mx + c 代入圆的方程并仔细化简。
- Use the discriminant only after writing the quadratic in standard form ax² + bx + c = 0.
- 只有在将二次方程化为标准形式 ax² + bx + c = 0 后才能使用判别式。
8. Using the Discriminant for Tangency | 利用判别式判断相切
A line is tangent to a circle if and only if the discriminant of the resulting quadratic is zero. This condition allows you to find unknown parameters in the line or circle equation.
当且仅当所得二次方程的判别式为零时，直线与圆相切。利用这一条件可以求出直线或圆方程中的未知参数。
et (1 + m²)x² + (-4 + 6m)x + 0 = 0. Set Δ = (-4 + 6m)² - 4(1 + m²)(0) = 0, giving m = 2/3.
例题：求 m 使 y = mx + 3 与 x² + y² - 4x - 6y + 9 = 0 相切。代入并化简得到 (1 + m²)x² + (-4 + 6m)x + 0 = 0。令 Δ = (-4 + 6m)² - 4(1 + m²)(0) = 0，解得 m = 2/3。
- If the quadratic has no x² term, the line may be vertical or horizontal; handle these separately.
- 如果二次方程没有 x² 项，直线可能是竖直或水平的，应单独处理。
- Always state that Δ = 0 gives a tangent, Δ > 0 gives two intersections, Δ < 0 gives none.
- 解题时务必写明：Δ = 0 为相切，Δ > 0 为两个交点，Δ < 0 为无交点。
9. Circles Through Given Points | 过给定点的圆
quation x² + y² + 2gx + 2fy + c = 0 to create three simultaneous equations in g, f, and c.
求经过三个给定点的圆的方程时，将每个点代入一般方程 x² + y² + 2gx + 2fy + c = 0，得到关于 g、f 和 c 的三个联立方程。
Solve the system to find g, f, and c, then write the circle equation. The centre and radius can then be determined using the formulas.
解这个方程组求出 g、f 和 c，然后写出圆的方程。圆心和半径可随后用公式求出。
- Alternatively, use the perpendicular bisector method with two chords.
- 也可以使用两条弦的垂直平分线方法。
- Check that all three points satisfy the final equation.
- 最后应检查三个点是否都满足所求方程。
10. Exam-Style Mixed Problems | 考试风格综合题
Exercise 6I often includes multi-step problems combining circle properties with straight lines, distances, and algebraic manipulation. Work systematically: draw a sketch, label known values, and write down the relevant geometric property before calculating.
练习 6I 常包含将圆的性质与直线、距离和代数运算相结合的多步骤问题。解题时应系统地处理：画出草图，标出已知量，并在计算前写出相关的几何性质。
Common tasks include proving a line is tangent, finding the equation of a tangent from an external point, and determining the length of a chord.
常见题型包括：证明一条直线是切线、求从圆外一点所作的切线方程，以及确定弦长。
- For tangent from external point, use the fact that the radius to the tangent point is perpendicular to the tangent, combined with the distance from the centre to the tangent line equal to r.
- 求圆外一点到圆的切线时，利用“圆心到切线的距离等于半径”以及“半径垂直于切线”的性质。
- For chord length, drop a perpendicular from the centre to the chord and use Pythagoras.
- 求弦长时，从圆心向弦作垂线，并用勾股定理。
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