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P2 2024 Jan Q3

A Level / Edexcel / P2

IAL 2024 Jan Paper · Question 3

题目

Problem

The circle CC

  • has centre A(3,5)A(3,5)
  • passes through the point B(8,−7)B(8,-7)

(a) Find an equation for CC.

(3)

The points MM and NN lie on CC such that MNMN is a chord of CC.

Given that MNMN

  • lies above the xx-axis
  • is parallel to the xx-axis
  • has length 4224\sqrt{22}

(b) find an equation for the line passing through points MM and NN.

(3)
题目中文翻译

圆 CC:

  • 圆心为 A(3,5)A(3,5);
  • 经过点 B(8,−7)B(8,-7)。

(a) 求 CC 的方程。

(3分)

点 MM 和 NN 在圆 CC 上,MNMN 是 CC 的一条弦。已知 MNMN:

  • 位于 xx 轴上方;
  • 平行于 xx 轴;
  • 长度为 4224\sqrt{22}。

(b) 求经过 MM、NN 的直线方程。

(3分)

解答

(a)

解法一

思路

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圆心已经给出,所以只需要用点 BB 求半径。圆的标准形式是 (x−a)2+(y−b)2=r2(x-a)^2+(y-b)^2=r^2。

答题过程

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The centre is A(3,5)A(3,5).

The radius is ABAB, so

r2= (8−3)2+(−7−5)2= 52+(−12)2= 25+144= 169.\begin{align*} r^2=&\,(8-3)^2+(-7-5)^2\\ =&\,5^2+(-12)^2\\ =&\,25+144\\ =&\,169. \end{align*}

Therefore, an equation of CC is

(x−3)2+(y−5)2=169.\begin{align*} (x-3)^2+(y-5)^2=169. \end{align*}

展开 (a) 的标准圆方程,也可写成 x2+y2−6x−10y−135=0x^2+y^2-6x-10y-135=0。

(b)

解法一

思路

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水平弦的中点在圆心正上方或正下方。先求出圆心到弦的距离,再检查两个候选高度;不能仅凭“在 xx 轴上方”就认定弦在圆心上方。半弦长是 2222\sqrt{22},半径是 1313,用直角三角形求圆心到弦的距离。

答题过程

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The radius of the circle is

r=13.\begin{align*} r=13. \end{align*}

Since MNMN has length 4224\sqrt{22}, half of the chord has length

222.\begin{align*} 2\sqrt{22}. \end{align*}

Let the perpendicular distance from AA to the chord MNMN be dd.

Using Pythagoras’ theorem,

d2+(222)2= 132d2+88= 169d2= 81d= 9.\begin{align*} d^2+(2\sqrt{22})^2 =&\, 13^2\\ d^2+88 =&\, 169\\ d^2 =&\, 81\\ d =&\, 9. \end{align*}

The possible heights of the horizontal chord are 5+9=145+9=14 and 5−9=−45-9=-4. Since MNMN lies above the xx-axis, reject −4-4 and choose

5+9=14.\begin{align*} 5+9=14. \end{align*}

Therefore, the equation of the line through MM and NN is

y=14.\begin{align*} y=14. \end{align*}

解法二

思路

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由于 MNMN 平行于 xx 轴,设直线方程为 y=ky=k(且 k>0k>0)。因为 MNMN 的长度为 4224\sqrt{22},且圆的对称轴为 x=3x=3,所以 M,NM, N 的 xx 坐标为 3±2223 \pm 2\sqrt{22}。将其代入圆方程中可以直接求解 kk。

答题过程

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Since the chord MNMN is parallel to the xx-axis, its equation can be written as y=ky = k (where k>0k > 0).

The centre of the circle is A(3,5)A(3,5) and the length of MNMN is 4224\sqrt{22}, which means the xx-coordinates of the points MM and NN are symmetrically placed at:

x=3±222\begin{align*} x = 3 \pm 2\sqrt{22} \end{align*}

Substitute these coordinates into the equation of CC from part (a):

(x−3)2+(y−5)2= 169(3±222−3)2+(k−5)2= 169(222)2+(k−5)2= 16988+(k−5)2= 169(k−5)2= 81k−5= ±9.\begin{align*} (x-3)^2+(y-5)^2 =&\, 169\\ (3 \pm 2\sqrt{22}-3)^2+(k-5)^2 =&\, 169\\ (2\sqrt{22})^2+(k-5)^2 =&\, 169\\ 88+(k-5)^2 =&\, 169\\ (k-5)^2 =&\, 81\\ k-5 =&\, \pm 9. \end{align*}

This gives two possible values:

k=14ork=−4\begin{align*} k = 14 \quad\text{or}\quad k = -4 \end{align*}

Since the chord lies above the xx-axis, we must have k>0k > 0, so k=14k = 14.

Therefore, the equation of the line is y=14y = 14.

解法三

思路

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用三角形 MNAMNA 的余弦定理求底角的精确余弦,再求正弦,得到圆心到弦的距离。保留精确值,避免把 14 算成舍入的近似值。

答题过程

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Let β=∠MNA\beta=\angle MNA. Since AM=AN=13AM=AN=13 and MN=422MN=4\sqrt{22}, the cosine rule gives

cos⁡β= 132+(422)2−1322×13×422= 22213.\begin{align*} \cos\beta =&\,\frac{13^2+(4\sqrt{22})^2-13^2} {2\times13\times4\sqrt{22}}\\ =&\,\frac{2\sqrt{22}}{13}. \end{align*}

The angle is acute, so

sin⁡β=1−88169=913.\sin\beta =\sqrt{1-\frac{88}{169}} =\frac9{13}.

The perpendicular distance from AA to MNMN is d=13sin⁡β=9d=13\sin\beta=9. Hence the candidate heights are 5±95\pm9, namely 1414 and −4-4.

Only 1414 is above the xx-axis, so

y=14.\boxed{y=14}.