## FANDOM

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What the locus of this:

$|z-1|=2 |z-i|$

### Geometrical viewEdit

The figure represented on an Argand diagram by this equation is the set of points twice as far from 1 as from i.

### Intersection of two conesEdit

In x-y-z space, let cone A be given by $z^2=\frac{(x-1)^2+y^2}{4},$ and cone B be given by $z^2=x^2+(y-1)^2.$ The projection of the intersection of these two cones onto the x-y plane is exactly the solution to the student's question. But what about the three-dimensional intersection of two cones? Since its projection onto the x-y plane is a circle, it might be an ellipse. But it can't be an ellipse because it's not a plane figure; it's a kind of saddle shape, conforming to the surface of the larger cone, and wrapped around the body of the smaller cone. More research is needed in this wiki regarding the intersection of two parallel right circular cones.[improvement needed: elaborate]

But what about the three-dimensional intersection of two cones? Since its projection onto the x-y plane is a circle, it might be an ellipse. But it can't be an ellipse because it's not a plane figure; it's a kind of saddle shape, conforming to the surface of the larger cone, and wrapped around the body of the smaller cone.

## SolutionEdit

First, we'll do the specific solution, then a more general solution.

### Specific solutionEdit

Letting z=x+iy, we see that plotting $|z-1|=2 |z-i|$ on an argand diagram is the same as plotting $(x-1)^2 + y^2 = 4\left(x^2 + (y-1)^2\right)$ on the x-y plane. Expanding $(x-1)^2 + y^2 - 4\left(x^2 + (y-1)^2\right) = 0$ gives $-3x^2-2x-3y^2+8y-3 = 0.$ Multiply through by −3 to give you $9x^2+6x+9y^2-24y+9 = 0.$ Now complete the squares as

$9x^2+6x+9y^2-24y = -9+1+16$
$9x^2+6x+1 + 9y^2-24y+16 = 8$
$(3x+1)^2 + (3y-4)^2 = 8$

So the solution is a circle with center $(\frac{-1}{3},\frac{4}{3})$ and radius $\frac{\sqrt 8}{3}.$

### General solutionEdit

Now, generalize. Plot $|z-A| = k|z-B|.$ Letting $A=a+bi$ and $B=c+di,$

$(x-a)^2 + (y-b)^2 = k^2((x-c)^2+(y-d)^2)$

Rewrite this as

$k^2((x-c)^2+(y-d)^2) - (x-a)^2 - (y-b)^2 = 0$

and then expand it, giving

$- x^2 + k^2x^2 + 2ax - 2ck^2x - y^2 + k^2y^2 + 2by - 2dk^2y = a^2 + b^2 - c^2k^2 - d^2k^2 = 0,$

then gather $x^2, x, y^2, y$ terms on the left, constants on right:

$(k^2-1)x^2 - (2ck^2-2a)x + (k^2-1)y^2 - (2dk^2-2b)y = a^2 + b^2 - c^2k^2 - d^2k^2$

now multiply through by $(k^2-1)$ so that the terms resemble $P^2x^2-2PQ + R^2y^2-2RS...$

$(k^2-1)^2x^2 - 2(k^2-1)(ck^2-a)x + (k^2-1)^2y^2 - 2(k^2-1)(dk^2-b)y = a^2k^2 + b^2k^2 - c^2k^4 - d^2k^4 - a^2 - b^2 + c^2k^2 + d^2k^2$

Now complete the squares, letting $P,R=(k^2-1)$ and $Q=(ck^2-a)$ and $S=(dk^2-b)$:

$\begin{split} (k^2-1)^2x^2 - 2(k^2-1)(ck^2-a)x + (ck^2-a)^2 + (k^2-1)^2y^2 - 2(k^2-1)(dk^2-b)y + (dk^2-b)^2 = \\ \quad \quad (ck^2-a)^2 + (dk^2-b)^2 + a^2k^2 + b^2k^2 - c^2k^4 - d^2k^4 - a^2 - b^2 + c^2k^2 + d^2k^2\end{split}$
$((k^2-1)x-(ck^2-a))^2 + ((k^2-1)y-(dk^2-b))^2 = (ck^2-a)^2 + (dk^2-b)^2 + a^2k^2 + b^2k^2 - c^2k^4 - d^2k^4 - a^2 - b^2 + c^2k^2 + d^2k^2$
$((k^2-1)x-(ck^2-a))^2 + ((k^2-1)y-(dk^2-b))^2 = a^2k^2 - 2ack^2 + c^2k^2 + b^2k^2 - 2bdk^2 + d^2k^2$
$((k^2-1)x-(ck^2-a))^2 + ((k^2-1)y-(dk^2-b))^2 = k^2((a-c)^2+(b-d)^2)$

So the center of the circle is $((ck^2-a)/(k^2-1), (dk^2-b)/(k^2-1)),$ and the radius is $k \sqrt{(a-c)^2+(b-d)^2)}/(k^2-1),$ which is $k/(k^2-1)$ times the distance between the centers of the original circles.