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CIE 9709 2026 June Paper 12 Q8

A Level / CIE / P1

CIE 9709 2026 June Paper 12 Paper · Question 8

题目

Problem

A diameter of a circle has end-points at (−2,4)(-2,4) and (8,8)(8,8).

(a) Find an equation of the circle.

(4)

(b) The line with equation y=x+6y=x+6 intersects the circle at the points PP and QQ.

Find the exact distance PQPQ. Give your answer in the form aba\sqrt{b}, where aa and bb are both integers greater than 11.

(7)
题目中文翻译

一个圆的一条直径的两个端点为 (−2,4)(-2,4) 和 (8,8)(8,8)。

(a) 求该圆的方程。

(4)

(b) 直线 y=x+6y=x+6 与该圆交于点 PP 和 QQ。

求距离 PQPQ 的精确值。答案写成 aba\sqrt{b} 的形式,其中 aa 和 bb 都是大于 11 的整数。

(7)
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

直径的中点是圆心。用圆心到一个端点的距离平方写标准方程。

答题过程

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The centre is (3,6)(3,6), the midpoint of (−2,4)(-2,4) and (8,8)(8,8). The squared radius is (8−3)2+(8−6)2=29(8-3)^2+(8-6)^2=29, so the equation is

(x−3)2+(y−6)2=29.(x-3)^2+(y-6)^2=29.

(b)

解法一

思路

展开

把直线代入圆方程求两个交点,再用两点距离公式。

答题过程

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With y=x+6y=x+6, the circle equation becomes (x−3)2+x2=29(x-3)^2+x^2=29. Thus

2x2−6x−20=2(x+2)(x−5)=0.2x^2-6x-20=2(x+2)(x-5)=0.

The points are (−2,4)(-2,4) and (5,11)(5,11). Therefore their separation is

PQ=(5+2)2+(11−4)2=72.PQ=\sqrt{(5+2)^2+(11-4)^2}=7\sqrt2.