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CIE 9709 2025 June Paper 12 Q8

A Level / CIE / P1

CIE 9709 2025 June Paper 12 Paper · Question 8

题目

Problem

The diagram shows the circle with equation x2+y2−14x+8y+36=0x^2 + y^2 - 14x + 8y + 36 = 0 and the line y=−2y = -2. The line intersects the circle at the points AA and BB. The centre of the circle is CC.

(a) Find the coordinates of AA, BB and CC.

[3]

(b) Find the angle ACBACB in radians. Give your answer correct to 3 significant figures.

[2]

(c) The chord ABAB divides the circle into two segments.

Find the area of the larger segment.

[4]
题目中文翻译

图中显示了方程为 x2+y2−14x+8y+36=0x^2 + y^2 - 14x + 8y + 36 = 0 的圆和直线 y=−2y = -2。该直线与圆交于点 AA 和 BB。圆心为 CC。

(a) 求 AA、BB、CC 的坐标。

[3]

(b) 求角 ACBACB 的大小,单位为弧度。答案精确到 3 位有效数字。

[2]

(c) 弦 ABAB 将圆分成两个弓形。

求较大弓形的面积。

[4]

解答

(a)

解法一

思路

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对圆的方程配方即可读出圆心。把 y=−2y=-2 代入标准式求弦的两个端点。

答题过程

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Completing the squares gives (x−7)2+(y+4)2=29(x-7)^2+(y+4)^2=29, so C=(7,−4)C=(7,-4). When y=−2y=-2, (x−7)2=25(x-7)^2=25, giving x=2x=2 or 1212. From the diagram, A=(2,−2)A=(2,-2) and B=(12,−2)B=(12,-2).

(b)

解法一

思路

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圆心到弦中点的垂线平分圆心角。用半弦和圆心到弦的距离求半角。

答题过程

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The midpoint of ABAB is M=(7,−2)M=(7,-2), so AM=5AM=5 and CM=2CM=2. Hence

∠ACB=2tan⁡−1(5/2)=2.38\angle ACB=2\tan^{-1}(5/2)=2.38

radians to three significant figures.

解法二

思路

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也可以直接用三角形 ACBACB 的三边与余弦定理。

答题过程

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The sides are AC=BC=29AC=BC=\sqrt{29} and AB=10AB=10. For ϕ=∠ACB\phi=\angle ACB,

cos⁡ϕ=29+29−1002(29)=−2129.\cos\phi=\frac{29+29-100}{2(29)}=-\frac{21}{29}.

This gives ϕ=2.38\phi=2.38 radians to three significant figures.

(c)

解法一

思路

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大弓形等于整圆减去小弓形。小弓形是小扇形减三角形 ACBACB。

答题过程

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Let ϕ\phi be the unrounded angle from (b). The minor sector area is 292ϕ\frac{29}{2}\phi, and the triangle with base 1010 and height 22 has area 1010. The larger segment therefore has area

29π−(292ϕ−10)≈66.629\pi-\Big(\frac{29}{2}\phi-10\Big)\approx66.6

square units.