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CIE 9709 2026 March Paper 12 Q6

A Level / CIE / P1

CIE 9709 2026 March Paper 12 Paper · Question 6

题目

Problem

The equation of a circle is

x2+y2=454.x^2+y^2=\frac{45}{4}.

(a) State the centre and radius of the circle.

(2)

(b) The line x+y=32x+y=\frac{3}{2} intersects the circle at the points AA and BB.

Find the exact distance ABAB.

(6)
题目中文翻译

一个圆的方程为

x2+y2=454.x^2+y^2=\frac{45}{4}.

(a) 写出该圆的圆心和半径。

(2)

(b) 直线 x+y=32x+y=\frac{3}{2} 与该圆交于点 AA 和 BB。

求距离 ABAB 的精确值。

(6)
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

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和圆的标准式比较。右边是半径的平方。

答题过程

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The centre is (0,0)(0,0) and the radius is 45/4=35/2\sqrt{45/4}=3\sqrt5/2.

(b)

解法一

思路

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把直线代入圆求两个端点,再用两点距离公式。

答题过程

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Substitute y=3/2−xy=3/2-x in the circle equation. This gives

2x2−3x−9=(2x+3)(x−3)=0.2x^2-3x-9=(2x+3)(x-3)=0.

The endpoints are (−3/2,3)(-3/2,3) and (3,−3/2)(3,-3/2). Each coordinate changes by 9/29/2 in magnitude, so the exact length is AB=92/2AB=9\sqrt2/2.

解法二

思路

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也可以先找圆心到弦的垂足。垂足平分弦,半径和圆心到弦的距离就能确定半弦长。

答题过程

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The perpendicular from the origin to x+y=3/2x+y=3/2 is y=xy=x, meeting the chord at M=(3/4,3/4)M=(3/4,3/4). Hence OM2=9/8OM^2=9/8. The chord length is

2454−98=922.2\sqrt{\frac{45}{4}-\frac98}=\frac{9\sqrt2}{2}.