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CIE 9709 2024 June Paper 12 Q7

A Level / CIE / P1

CIE 9709 2024 June Paper 12 Paper · Question 7

题目

Problem

The equation of a circle is (x−6)2+(y+a)2=18(x - 6)^2 + (y + a)^2 = 18. The line with equation y=2a−xy = 2a - x is a tangent to the circle.

(a) Find the two possible values of the constant aa.

[5]

(b) For the greater value of aa, find the equation of the diameter which is perpendicular to the given tangent.

[3]
题目中文翻译

一个圆的方程为 (x−6)2+(y+a)2=18(x - 6)^2 + (y + a)^2 = 18。方程为 y=2a−xy = 2a - x 的直线是该圆的一条切线。

(a) 求常数 aa 的两个可能值。

[5]

(b) 对于较大的 aa 值,求垂直于给定切线的直径的方程。

[3]

解答

(a)

解法一

思路

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把切线代入圆的方程。相切意味着联立所得的二次方程恰好有一个实根。

答题过程

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Substituting y=2a−xy=2a-x gives (x−6)2+(3a−x)2=18(x-6)^2+(3a-x)^2=18, which expands to

2x2−(12+6a)x+9a2+18=0.2x^2-(12+6a)x+9a^2+18=0.

The discriminant is (12+6a)2−8(9a2+18)=−36a2+144a(12+6a)^2-8(9a^2+18)=-36a^2+144a. Setting it to zero gives a(a−4)=0a(a-4)=0. Thus a=0a=0 or a=4a=4.

解法二

思路

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也可用圆心到切线的距离等于半径。直线的法向量长度用于把代入值转换成距离。

答题过程

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The centre is (6,−a)(6,-a) and the radius is 18\sqrt{18}. The distance to x+y−2a=0x+y-2a=0 is

∣6−3a∣2.\frac{|6-3a|}{\sqrt2}.

For tangency this equals 18\sqrt{18}, so ∣6−3a∣=6|6-3a|=6. Hence a=0a=0 or a=4a=4.

(b)

解法一

思路

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直径经过圆心,且与斜率为 −1-1 的切线垂直。

答题过程

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With a=4a=4, the centre is (6,−4)(6,-4). The diameter has gradient 11, so its equation is y+4=x−6y+4=x-6, or y=x−10y=x-10.