Solving $|z-4i|=2|z+4|$

Question

$$\begin{align} |z-4i|&=2|z+4|\\[4pt] |x+yi-4i|&=2|x+yi+4|\\[4pt] |x+i(y-4)|&=2|(x+4)+iy|\\[4pt] \sqrt{x^2+(y-4)^2}&=2\sqrt{(x+4)^2+y^2}\\[4pt] (\sqrt{x^2+(y-4)^2})^2&=(2\sqrt{(x+4)^2+y^2})^2\\[4pt] x^2+y^2-8y+16&= 4(x^2+8x+16+y^2)\\[4pt] x^2+y^2-8y+16&= 4x^2+32x+64+4y^2\\[4pt] 0&= 3x^2+3y^2+32x+8y+48 \end{align}$$

Is it okay? Thank you

Answer 1

Your computation is fine.

But I’ve been meditating on the form of your final result, since it says that the locus is a circle: coefficients of $x^2$ and $y^2$ are the same, and no $xy$ term. So there must be a theorem from plane geometry that says: If $P$ and $Q$ are points in the plane, the locus of points $R$ that are $k$ times as far from $Q$ as from $P$ is a circle, if $k\ne1$.

It’s easy enough to prove this analytically, by a method the same as yours, but surely there has to be a synthetic proof, perhaps even straight out of Euclid.

Does anyone know such a proof? Or can you cook one up? I looked at it, and have been stumped so far.

Anyway, thanks for a provocative question (and calculation).

Answer 2

So far so good.

You may continue to see the geometry of the solution as well.

$$3x^2+3y^2+32x+8y+48=0$$

$$x^2+y^2+(32/3)x+(8/3)y+16=0$$

Which is the equation of a circle.

Answer 3

Your solution looks fine, but you should realize that it is an equation for the circle with center $\left(-\dfrac{16}{3},-\dfrac{4}{3}\right)$ and with radius $\dfrac{8\sqrt{2}}{3}$. If I were your grader, I would not give you a full credit for simply finding the final equation, yet not realizing it gives a circle. Here is an alternative solution, using Euclidean geometry of the plane.

Let $A$ denote the point $4\text{i}$ of the complex plane $\mathbb{C}\cong\mathbb{R}^2$, whilst $B$ is the point $-4$. Thus, if the point $C$ with complex coordinate $z$ satisfies $$|z-4\text{i}|=2\,|z+4|\,,$$ then $$CA=2\,CB\,.$$ On the line $AB$, there are two solutions $D$ and $E$, with complex coordinates $$\frac{1}{3}\,(4\text{i})+\frac{2}{3}\,(-4)=\frac{-8+4\text{i}}{3}\text{ and }(-1)\,(4\text{i})+2\,(-4)=-8-4\text{i}\,,$$ respectively. Thus, the point $C$ is a point such that $CD$ is the internal angular bisector of $\angle ACB$ and $CE$ is the external angular bisector of $\angle ACB$. We can easily show that the locus of $C$ is a circle $\Gamma$ with diameter $DE$.

Thus, the center $P$ of $\Gamma$ has the complex coordinate $$\frac{1}{2}\,\left(\frac{-8+4\text{i}}{3}\right)+\frac{1}{2}\,(-8-4\text{i})=\frac{-16-4\text{i}}{3}\,.$$ The radius of $\Gamma$ is $$\frac{1}{2}\,\Biggl|\left(\frac{-8+4\text{i}}{3}\right)-(-8-4\text{i})\Biggr|=\frac{8\sqrt{2}}{3}\,.$$ In other words, the complex coordinate $z$ of $C$ satisfies $$\Biggl|z-\left(\frac{-16-4\text{i}}{3}\right)\Biggr|=\frac{8\sqrt{2}}{3}\,.$$ We can also write $$z=\left(\frac{-16-4\text{i}}{3}\right)+\frac{8\sqrt{2}}{3}\,\exp(\text{i}\theta)\,,$$ where $\theta\in\mathbb{R}$.

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In general, any solution $z\in\mathbb{C}$ to $|z-a|=r\,|z-b|$, where $r\in\mathbb{R}_{>0}\setminus\{1\}$ and $a,b\in\mathbb{C}$ is given by the circle $$\left|z-c\right|=\rho\,.$$ Here, $c:=\dfrac{-a+r^2b}{r^2-1}$ and $\rho:=\dfrac{r}{|r^2-1|}\,|a-b|$. In other words, $$z=c+\rho\,\exp(\text{i}\theta)\,,$$ where $\theta\in\mathbb{R}$. (Note that the solutions in the case where $r=1$ form a degenerate circle---a straight line. This straight line is the perpendicular bisector of the segment joining $a$ and $b$. It is given by the equation $(a-b)\,\bar{z}+(\bar{a}-\bar{b})z=|a|^2-|b|^2$, or equivalently, $\text{Re}\big((\bar{a}-\bar{b})\,z\big)=\dfrac{|a|^2-|b|^2}{2}$. In other words, $z=\dfrac{a+b}{2}+\text{i}(a-b)\,t$, where $t\in\mathbb{R}$.)

Answer 4

This is the set of points which are twice as far from $-4$ as from $4i$.

I'll work out the case of arbitrary points and ratio of distances.

If the points are $(a, b)$ and $(c, d)$ and the ratio is $r$, then $|z-(a, b)| = r|z-(c, d)|$ or, if $z = (x, y)$, $\sqrt{(x-a)^2+(y-b)^2} =r\sqrt{(x-c)^2+(y-d)^2} $.

Squaring and expanding, $x^2-2ax+a^2+y^2-2by+b^2 =r^2(x^2-2cx+c^2+y^2-2dy+d^2) $.

Grouping, $x^2(r^2-1)+2(a-r^2c)x+y^2(r^2-1)+2(b-r^2d)y +r^2(c^2+d^2)-a^2-b^2 =0 $.

If $r=1$, this becomes $2(a-c)x+2(b-d)y +c^2+d^2-a^2-b^2 =0 $, which is the equation of a straight line. This is because the set of points equidistant from two points is the perpendicular bisector of the line joining the two points.

If $r \ne 1$, this is of the form $Ax^2+Bx+Ay^2+Cy+D =0 $ which is the equation of a circle. To see this, write this as

$\begin{array}\\ 0 &=Ax^2+Bx+Ay^2+Cy+D\\ &=A\left(x^2+(B/A)x+y^2+(C/A)y+D/A\right)\\ &=A\left(x^2+(B/A)x+(B^2/4A^2)-(B^2/4A^2)+y^2+(C/A)y+(C^2/4A^2)-(C^2/4A^2)+D/A\right)\\ &=A\left((x+B/2A)^2+(y+C/2A)^2+D/A-(B^2/4A^2)-(C^2/4A^2)\right)\\ &=A\left((x+B/2A)^2+(y+C/2A)^2+(4AD-B^2-C^2)/(4A^2)\right)\\ \end{array} $

or $(x+B/2A)^2+(y+C/2A)^2 =(B^2+C^2-4AD)/(4A^2) $.

This is a circle with center at $-B/2A, -C/2A)$ and radius $\sqrt{B^2+C^2-4AD}/(2A) $. For the circle to have a real radius, we must have $B^2+C^2-4AD \ge 0$.

In our case,

$\begin{array}\\ B^2+C^2-4AD &=(2(a-r^2c))^2+(2(b-r^2d))^2-4(r^2-1)(r^2(c^2+d^2)-a^2-b^2)\\ &=4\left(a^2-2r^2ac+r^4c^2+b^2-2r^2bd+r^4d^2-(r^2-1)r^2(c^2+d^2)+(r^2-1)(a^2+b^2)\right)\\ &=4\left((a^2+b^2)(1+(r^2-1))+(c^2+d^2)(1-r^2(r^2-1))-2ac-2bd)\right)\\ &=4\left(r^2(a^2+b^2)+(c^2+d^2)(r^4-r^2(r^2-1))-2r^2(ac+bd)\right)\\ &=4\left(r^2(a^2+b^2)+r^2(c^2+d^2)-2r^2(ac+bd)\right)\\ &=4r^2\left((a^2+b^2)+(c^2+d^2)-2(ac+bd)\right)\\ &=4r^2\left((a^2-2ac+c^2)+(b^2-2bd+d^2)\right)\\ &=4r^2\left((a-c)^2+(b-d)^2)\right)\\ &\ge 0\\ \end{array} $

The radius is zero if and only if $a=c$ and $b=d$, which means the two points are the same.

Otherwise the radius is $\dfrac{2r\sqrt{(a-c)^2+(b-d)^2}}{2(r^2-1)} =\dfrac{r\sqrt{(a-c)^2+(b-d)^2}}{r^2-1} $.

The center is at $ (-\dfrac{a-r^2c}{r^2-1}, -\dfrac{b-r^2d}{r^2-1}) =(\dfrac{r^2c-a}{r^2-1}, \dfrac{r^2d-b}{r^2-1}) =\dfrac{1}{r^2-1}(r^2(c, d)-(a, b)) $.

Note that if $r=0$ then the center is at $(a, b)$ and the radius is zero, so the circle is a point at $(a, b)$.

Similarly, if $r\to \infty$, then the center goes to $(c, d)$ and the radius goes to zero, so the circle again is a point.

Answer 5

Here's another approach altogether:

$$\begin{align} |z-4i|=2|z+4| &\iff\left|z-4i\over z+4 \right|=2\\ &\iff{z-4i\over z+4 }=2e^{i\theta}\quad\text{for some }\theta\in\mathbb{R}\\ &\iff z={8e^{i\theta}+4i\over1-2e^{i\theta}} \end{align}$$

Answer 6

Your argument is completely fine. Continue it with: $$x^2+\frac{32}{3}x+y^2+\frac 83 y+16=0$$ Now complete the square on the $x$'s and $y$'s $$(x+\frac{16}{3})^2-\frac{256}{9}+(y+\frac43)^2-\frac{16}{9}+16=0$$

$$(x+\frac{16}{3})^2+(y+\frac{4}{3})^2=\frac{128}{9}$$ Thus we have a circle, centre $(-\frac{16}{3}, -\frac 43)$, radius $\frac{8\sqrt2}{3}$.