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CIE 9709 2026 June Paper 15 Q11

A Level / CIE / P1

CIE 9709 2026 June Paper 15 Paper · Question 11

题目

Problem

In the diagram, ABAB is a chord of the circle with equation (x−6)2+(y−3)2=20(x-6)^2+(y-3)^2=20. The two tangents to the circle with equations y=2x+1y=2x+1 and y=−2x+25y=-2x+25 meet the circle at the points AA and BB respectively.

(a) Find the coordinates of the points AA and BB.

(4)

(b) Hence, or otherwise, find the area of the shaded region. Give your answer to 33 significant figures.

(5)
题目中文翻译

图中,ABAB 是圆 (x−6)2+(y−3)2=20(x-6)^2+(y-3)^2=20 的一条弦。该圆的两条切线 y=2x+1y=2x+1 和 y=−2x+25y=-2x+25 分别在点 AA 和 BB 处与圆相切。

(a) 求点 AA 和 BB 的坐标。

(4)

(b) 由此或用其他方法,求阴影区域的面积。答案保留 33 位有效数字。

(5)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。

解答

(a)

解法一

思路

展开

每条切线都与过切点的半径垂直。由圆心和负倒数斜率写出半径方程,再分别求交。

答题过程

展开

The centre is C=(6,3)C=(6,3). Since the tangent at AA has gradient 22, the radius line CACA has equation y=6−x/2y=6-x/2. Intersecting it with y=2x+1y=2x+1 gives A=(2,5)A=(2,5). The tangent at BB has gradient −2-2, so the radius line CBCB is y=x/2y=x/2. Intersecting it with y=−2x+25y=-2x+25 gives B=(10,5)B=(10,5).

解法二

思路

展开

也可以把切线方程代入圆。切线与圆只有一个交点,二次式会出现重根。

答题过程

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Substituting y=2x+1y=2x+1 in the circle equation gives 5(x−2)2=05(x-2)^2=0, hence x=2x=2 and y=5y=5. Substituting y=−2x+25y=-2x+25 gives 5(x−10)2=05(x-10)^2=0, hence x=10x=10 and y=5y=5. Thus A=(2,5)A=(2,5) and B=(10,5)B=(10,5).

(b)

解法一

思路

展开

弦的中点把圆心角平分。由半弦长和圆心到弦的距离求圆心角,再用扇形面积减三角形面积。

答题过程

展开

The chord midpoint is M=(6,5)M=(6,5). Hence CM=2CM=2, AM=4AM=4, and the central angle is ϕ=2arctan⁡(4/2)=2arctan⁡2\phi=2\arctan(4/2)=2\arctan2 radians. The shaded segment has area

12(20)ϕ−12(8)(2)=20arctan⁡2−8≈14.1\frac12(20)\phi-\frac12(8)(2)=20\arctan2-8\approx14.1

to three significant figures.