Consider the following: $|z-1| = 3|z-2|$ where $z \in \mathbb{C}$. Geometrically, is this a circle with center $(\frac{17}{8}, 0)$ and radius $\frac{3}{8}$? If so, is their a relatively easy way to find this? The way I did it was squaring everything, then solving it until I had an equation of a circle. It was quite a bit of work, so I did not know if their was some more simple way to do this.
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$\begingroup$ Using elementary equivalences you can rewrite this as $\frac{1}{3} = |1 - \frac{1}{z-1}|$, and this defines a circle by general principles - inversion of a circle. But maybe this is not simpler. $\endgroup$– CocopuffsAug 27, 2013 at 5:18
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2 Answers
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Another way to see this is using Möbius transformations. Let $\phi(z)=\frac{z-1}{z-2}$ be a Möbius transformation. we know that these transformations map circles to circles and that composition of Möbius transformations is done like matrix multiplication, hence the inverse of $\phi$ is easy to find: $\phi^{-1}(z)=\frac{2z-1}{z-1}$. Then all you need to do is compute the image of $|z|=3$ under $\phi^{-1}$. (actually, I don't think this method is better for computation, but it makes the fact that it turns out to be a circle quite trivial)
To see why Möbius transformations maps circles to circles, I really recommend viewing this .
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A way to interpret this equation is: all points such that they are three times farther from $1$ than $2$. If you happen to know about Apollonian circles, this shows it is a circle.
To find the radius and center, find points on the line through $1$ and $2$. There is a point between them, $7/4$, and a point on the right side, $5/2$. Since this is a diameter of the line, the radius is $3/8$, and the center is just the average of the two points: $17/8$.
answered Aug 27, 2013 at 5:16