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CIE 9709 2024 March Paper 12 Q10

A Level / CIE / P1

CIE 9709 2024 March Paper 12 Paper · Question 10

题目

Problem

The diagram shows the circle with centre C(−4,5)C(-4, 5) and radius 20\sqrt{20} units. The circle intersects the yy-axis at the points AA and BB. The size of angle ACBACB is θ\theta radians.

(a) Find the equation of the tangent to the circle at the point (−6,9)(-6, 9).

[3]

(b) Find the equation of the circle in the form x2+y2+ax+by+c=0x^2 + y^2 + ax + by + c = 0.

[2]

(c) Find the value of θ\theta correct to 4 significant figures.

[3]

(d) Find the perimeter and area of the segment shaded in the diagram.

[4]
题目中文翻译

图中显示了圆心为 C(−4,5)C(-4, 5)、半径为 20\sqrt{20} 个单位的圆。该圆与 yy 轴相交于点 AA 和 BB。角 ACBACB 的大小为 θ\theta 弧度。

(a) 求该圆在点 (−6,9)(-6, 9) 处的切线方程。

[3]

(b) 求该圆的方程,并写成 x2+y2+ax+by+c=0x^2 + y^2 + ax + by + c = 0 的形式。

[2]

(c) 求 θ\theta 的值,精确到 4 位有效数字。

[3]

(d) 求图中阴影弓形的周长和面积。

[4]

解答

(a)

解法一

思路

展开

圆心到切点的连线与切线垂直。先算半径的斜率,再取负倒数。

答题过程

展开

The radius from (−4,5)(-4,5) to (−6,9)(-6,9) has slope (9−5)/(−6+4)=−2(9-5)/(-6+4)=-2. The tangent slope is 1/21/2, so its equation is y−9=12(x+6)y-9=\frac12(x+6), or y=12x+12y=\frac12x+12.

(b)

解法一

思路

展开

把圆心和半径代入圆的标准式,再展开并把常数移到左边。

答题过程

展开

The equation (x+4)2+(y−5)2=20(x+4)^2+(y-5)^2=20 expands to

x2+y2+8x−10y+21=0.x^2+y^2+8x-10y+21=0.

(c)

解法一

思路

展开

弦 ABAB 在 yy 轴上,弦中点和圆心组成一个直角三角形。用半弦长除以圆心到弦的距离求半角。

答题过程

展开

When x=0x=0, (y−5)2=4(y-5)^2=4, so the intercepts have y=3y=3 and y=7y=7. Their midpoint is (0,5)(0,5), which is 44 units from the centre, and the half-chord length is 22. Thus

θ=2tan⁡−1(2/4)=0.9273\theta=2\tan^{-1}(2/4)=0.9273

radians to four significant figures.

解法二

思路

展开

也可以用三边直接求圆心角。两腰都是半径,底边是两个纵截距的差。

答题过程

展开

In triangle ACBACB, the sides are 20\sqrt{20}, 20\sqrt{20} and 44. Thus

cos⁡θ=20+20−422(20)=35,\cos\theta=\frac{20+20-4^2}{2(20)}=\frac35,

giving θ=0.9273\theta=0.9273 radians to four significant figures.

(d)

解法一

思路

展开

周长由弦和弧组成。面积由扇形减去三角形得到,角度使用 (c) 的未舍入值。

答题过程

展开

The perimeter is 4+20 θ≈8.154+\sqrt{20}\,\theta\approx8.15 units. The sector area is 12(20)θ=10θ\frac12(20)\theta=10\theta, and the triangle area is 12(4)(4)=8\frac12(4)(4)=8. Thus the shaded area is 10θ−8≈1.2710\theta-8\approx1.27 square units.