Identify the plane defined by $|z-2i| = 2|z+3|$

Question

I tried:

$$|z-2i| = 2|z+3| \Leftrightarrow \\ |x+yi-2i|=2|x+yi+2|\Leftrightarrow \\ \sqrt{x^2+(y-2)^2}=\sqrt{4((x+2)^2+y^2)} \Leftrightarrow \\ \sqrt{x^2+y^2-4y+4} = \sqrt{4x^2+24x+36+4y^2} \Leftrightarrow \\ x^2+y^2-4y+4 = 4x^2+24x+36+4y^2 \Leftrightarrow \\ y^2-4y-4y^2=4x^2+24x+36+x^2 \Leftrightarrow \\ -3y^2-4y=5x^2+24x+26 \Leftrightarrow \\ ???$$

What do I do next?

It looks like you made a mistake going from the third to fourth lines, on the right. — B. Mehta, May 14 '17 at 01:28
You must mean the *locus* of $z$, not plane. The locus of points with constant ratio between distances to two fixed points is a circle. See for example here for the complete derivation. — dxiv, May 14 '17 at 01:36

bigfocalchord · Accepted Answer · 2017-05-15T19:48:23.260

3

I assume this means locus and not plane..

$$ \begin{align*} |z-2i|&= 2|z+3|\\ |x+iy-2i|&= 2|x+iy+3| \\ |x+i(y-2)|&= 2|(x+3)+i(y)| \\ \sqrt{x^2+(y-2)^2}&=2\sqrt{(x+3)^2+y^2}\\ x^2+(y-2)^2&=4\left(x^2+6x+9+y^2\right)\\ x^2+y^2-4y+4&=4x^2+24x+36+4y^2 \\ 3x^2+24x+32+3y^2+4y&=0\\ x^2+8x+\frac{32}{3}+y^2+\frac{4}{3}y&=0 \\ x^2+8x+y^2+\frac{4}{3}y &= -\frac{32}{3}\\ x^2+8x+16 +y^2+\frac{4}{3}y +\frac{4}{9}&= -\frac{32}{3}+16+\frac{4}{9} \\ (x+4)^2+\left(y+\frac{2}{3}\right)^2&=\frac{52}{9} \end{align*}$$

Which means our locus is a circle with centre $\left(-4,\frac{-2}{3}\right)$ and radius $\frac{2\sqrt{13}}{3}$.

edited May 15 '17 at 19:48

answered May 14 '17 at 01:45

bigfocalchord

5,597

Could you explain your last two steps? What did you do exactly? – Mark Read May 14 '17 at 15:21
@MarkRead have you heard of completing the square? – bigfocalchord May 14 '17 at 19:34
I have, but I don't use it much and I don't understand how you applied it in this case. Did you use it separately for x and y? – Mark Read May 14 '17 at 23:21
@MarkRead Yeah complete the square for both x,y seperately – bigfocalchord May 14 '17 at 23:31
I don't understand how completing the square is applicable here though. Shouldn't it be used only when you got an equation of the type $ax^2 + bx + c = 0$? Here you got two variables, how does that work? – Mark Read May 15 '17 at 02:03
And by the way, why did you divide by 3? – Mark Read May 15 '17 at 02:16
@MarkRead I added my completing the square steps – bigfocalchord May 15 '17 at 07:53
I see, thanks.. – Mark Read May 15 '17 at 16:39

score 2 · Answer 2 · answered May 15 '17 at 20:24

You can square without adding spurious solutions, because $|w|\ge0$ by definition. So $|z-2i|^2=4|z+3|^2$ and, using $|w|^2=w\bar{w}$, $$ (z-2i)(\bar{z}+2i)=4(z+3)(\bar{z}+3) $$ This simplifies to $$ 3z\bar{z}+12(z+\bar{z})-2i(z-\bar{z})+32=0 $$ Now recall that, for $z=x+yi$, $z\bar{z}=x^2+y^2$, $z+\bar{z}=2x$ and $z-\bar{z}=2yi$, to get $$ 3(x^2+y^2)+24x+4y+32=0 $$ that can also be written as $$ x^2+y^2+8x+\frac{4}{3}y+\frac{32}{3}=0 $$ or, completing the squares, $$ (x+4)^2+\left(y+\frac{2}{3}\right)^{\!2}=\frac{52}{9} $$ which represents a circle.

Your computation is also good up to $$ x^2+y^2-4y+4 = 4x^2+24x+36+4y^2 $$ Then you do wrong simplifications: taking everything to the right hand side you get $$ 3x^2+3y^2+24x+4y+32=0 $$ exactly the same I got.

score 1 · Answer 3 · answered May 20 '17 at 03:41

Naive geometric solution:

The locus of points whose distances are a fixed ratio (not equality) from two given points is a circle. Call the points $A$ and $B$. The circle's center lies on the (extended) line $AB$.

If you want all the points twice as far from $A$ as from $B$, then the two points where its circumference intersects line $AB$ are $\frac13$ of the way from $B$ to $A$, and at distance $d(AB)$ on the other side of $B$ (opposite from $A$). The center is the midpoint between these points, at distance $\frac13 d(AB)$ on the far side of $B$ from $A$. That makes the radius of the circle $\frac{2}{3}d(AB)$.

In this case, $A$ is at $2i$ and $B$ is at $-3$. That puts the center at the point $2i(1-t)-3t$ for $t=\frac{4}{3}$, and it makes the radius $\frac{2}{3}|2i+3|$. I.e., the center is at $-4-\frac23i$, and the radius is $\frac{2\sqrt{13}}{3}$

Identify the plane defined by $|z-2i| = 2|z+3|$

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