CalcGospel 國際數學圖譜
返回

FP2 2021 Oct Q6

A Level / Edexcel / FP2

IAL 2021 Oct Paper · Question 6

题目

Problem

6. The complex number zz on an Argand diagram is represented by the point PP where

 ∣z+1−13i∣= 3∣z−7−5i∣.\begin{align*} &\,|z+1-13\mathrm{i}|\\[4mm] =&\,3|z-7-5\mathrm{i}|. \end{align*}

Given that the locus of PP is a circle,

(a) determine the centre and radius of this circle.

(5)

The complex number ww, on the same Argand diagram, is represented by the point QQ, where

 arg⁡(w−8−6i)= −3π4.\begin{align*} &\,\arg(w-8-6\mathrm{i})\\[4mm] =&\,-\frac{3\pi}{4}. \end{align*}

Given that the locus of PP intersects the locus of QQ at the point RR,

(b) determine the complex number representing RR.

(4)
题目中文翻译

6. 复数 zz 在 Argand 图上由点 PP 表示，满足

 ∣z+1−13i∣= 3∣z−7−5i∣.\begin{align*} &\,|z+1-13\mathrm{i}|\\[4mm] =&\,3|z-7-5\mathrm{i}|. \end{align*}

已知 PP 的轨迹是一个圆。

(a) 求此圆的圆心和半径。

(5)

复数 ww 在同一 Argand 图上由点 QQ 表示，满足

 arg⁡(w−8−6i)= −3π4.\begin{align*} &\,\arg(w-8-6\mathrm{i})\\[4mm] =&\,-\frac{3\pi}{4}. \end{align*}

已知 PP 的轨迹与 QQ 的轨迹相交于点 RR。

(b) 求表示点 RR 的复数。

(4)

解答

(a)

解法一

思路

展开

令 z=x+iyz=x+\mathrm{i}y，把 modulus equation 转成 Cartesian equation。整理成圆的标准形式后，就能读出 centre 和 radius。

答题过程

展开

Let

z=x+iy.\begin{align*} z=x+\mathrm{i}y. \end{align*}

Then

 z+1−13i= (x+1)+i(y−13),\begin{align*} &\,z+1-13\mathrm{i}\\[4mm] =&\,(x+1)+\mathrm{i}(y-13), \end{align*}

and

 z−7−5i= (x−7)+i(y−5).\begin{align*} &\,z-7-5\mathrm{i}\\[4mm] =&\,(x-7)+\mathrm{i}(y-5). \end{align*}

So

 ∣z+1−13i∣= (x+1)2+(y−13)2,\begin{align*} &\,|z+1-13\mathrm{i}|\\[4mm] =&\,\sqrt{(x+1)^2+(y-13)^2}, \end{align*}

and

 ∣z−7−5i∣= (x−7)2+(y−5)2.\begin{align*} &\,|z-7-5\mathrm{i}|\\[4mm] =&\,\sqrt{(x-7)^2+(y-5)^2}. \end{align*}

The given equation becomes

 (x+1)2+(y−13)2= 3(x−7)2+(y−5)2.\begin{align*} &\,\sqrt{(x+1)^2+(y-13)^2}\\[4mm] =&\,3\sqrt{(x-7)^2+(y-5)^2}. \end{align*}

Square both sides:

 (x+1)2+(y−13)2= 9[(x−7)2 +(y−5)2].\begin{align*} &\,(x+1)^2+(y-13)^2\\[4mm] =&\,9\big[(x-7)^2\\[4mm] &\,\hspace{2pt}+(y-5)^2\big]. \end{align*}

Expand:

 x2+2x+1 +y2−26y+169= 9[x2−14x+49 +y2−10y+25]= 9x2−126x+441 +9y2−90y+225.\begin{align*} &\,x^2+2x+1\\[4mm] &\,\hspace{2pt}+y^2-26y+169\\[4mm] =&\,9\big[x^2-14x+49\\[4mm] &\,\hspace{2pt}+y^2-10y+25\big]\\[4mm] =&\,9x^2-126x+441\\[4mm] &\,\hspace{2pt}+9y^2-90y+225. \end{align*}

Bring all terms to one side and simplify:

0= 8x2+8y2−128x −64y+496.\begin{align*} 0=&\,8x^2+8y^2-128x\\[4mm] &\,\hspace{2pt}-64y+496. \end{align*}

Divide by 88:

 x2+y2−16x −8y+62= 0.\begin{align*} &\,x^2+y^2-16x\\[4mm] &\,\hspace{2pt}-8y+62\\[4mm] =&\,0. \end{align*}

Complete the square:

 x2−16x +y2−8y+62= 0, (x−8)2−64 +(y−4)2−16+62= 0.\begin{align*} &\,x^2-16x\\[4mm] &\,\hspace{2pt}+y^2-8y+62\\[4mm] =&\,0,\\[4mm] &\,(x-8)^2-64\\[4mm] &\,\hspace{2pt}+(y-4)^2-16+62\\[4mm] =&\,0. \end{align*}

So

(x−8)2+(y−4)2=18.\begin{align*} (x-8)^2+(y-4)^2=18. \end{align*}

Therefore the centre is

(8,4),\begin{align*} (8,4), \end{align*}

and the radius is

18=32.\begin{align*} \sqrt{18}=3\sqrt2. \end{align*}

(b)

解法一

思路

展开

由

arg⁡(w−8−6i)=−3π4\begin{align*} \arg(w-8-6\mathrm{i})=-\frac{3\pi}{4} \end{align*}

可知点 QQ 在从 (8,6)(8,6) 出发、方向角为 −3π4-\dfrac{3\pi}{4} 的射线上。这条射线所在直线斜率是 11，所以先写出直线方程，再与 (a) 的圆联立。

答题过程

展开

The locus

arg⁡(w−8−6i)=−3π4\begin{align*} \arg(w-8-6\mathrm{i})=-\frac{3\pi}{4} \end{align*}

is a half-line starting from (8,6)(8,6) with gradient

tan⁡(−3π4)=1.\begin{align*} \tan\bigg(-\frac{3\pi}{4}\bigg)=1. \end{align*}

So its Cartesian equation is

y−6=x−8.\begin{align*} y-6=x-8. \end{align*}

Hence

y=x−2.\begin{align*} y=x-2. \end{align*}

From part (a), the circle is

 x2+y2−16x −8y+62= 0.\begin{align*} &\,x^2+y^2-16x\\[4mm] &\,\hspace{2pt}-8y+62\\[4mm] =&\,0. \end{align*}

Substitute y=x−2y=x-2:

 x2+(x−2)2−16x −8(x−2)+62= 0.\begin{align*} &\,x^2+(x-2)^2-16x\\[4mm] &\,\hspace{2pt}-8(x-2)+62\\[4mm] =&\,0. \end{align*}

Expand:

 x2+x2−4x+4 −16x−8x+16+62= 0.\begin{align*} &\,x^2+x^2-4x+4\\[4mm] &\,\hspace{2pt}-16x-8x+16+62\\[4mm] =&\,0. \end{align*}

So

2x2−28x+82=0.\begin{align*} 2x^2-28x+82=0. \end{align*}

Divide by 22:

x2−14x+41=0.\begin{align*} x^2-14x+41=0. \end{align*}

Solve:

x= 14±196−1642= 14±422= 7±22.\begin{align*} x =&\,\frac{14\pm\sqrt{196-164}}{2}\\[2mm] =&\,\frac{14\pm4\sqrt2}{2}\\[2mm] =&\,7\pm2\sqrt2. \end{align*}

The half-line has direction −3π4-\dfrac{3\pi}{4} from (8,6)(8,6), so the point must lie down-left from (8,6)(8,6). Hence we choose

x=7−22.\begin{align*} x=7-2\sqrt2. \end{align*}

Then

y=x−2=5−22.\begin{align*} y=x-2=5-2\sqrt2. \end{align*}

Therefore the complex number representing RR is

(7−22)+(5−22)i.\begin{align*} (7-2\sqrt2)+(5-2\sqrt2)\mathrm{i}. \end{align*}