CalcGospel 國際數學圖譜
返回

IAL 2021 Oct Q6

A Level / Edexcel / FP2

IAL 2021 Oct Paper · Question 6

题目

Problem

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}| = 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}) = -\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)

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} =&\,(x+1)+\mathrm{i}(y-13), \end{align*}

and

z−7−5i= (x−7)+i(y−5).\begin{align*} z-7-5\mathrm{i} =&\,(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}| =&\,\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}| =&\,\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} =&\,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 =&\,9\left[(x-7)^2+(y-5)^2\right]. \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+y^2-26y+169 =&\,9\left(x^2-14x+49+y^2-10y+25\right)\\[2mm] =&\,9x^2-126x+441+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-64y+496. \end{align*}

Divide by 88:

x2+y2−16x−8y+62=0.\begin{align*} x^2+y^2-16x-8y+62=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+y^2-8y+62 =&\,0,\\[2mm] (x-8)^2-64+(y-4)^2-16+62 =&\,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\left(-\frac{3\pi}{4}\right)=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-8y+62=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-8(x-2)+62=0. \end{align*}

Expand:

x2+x2−4x+4−16x−8x+16+62=0.\begin{align*} x^2+x^2-4x+4-16x-8x+16+62=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*}