So, I basically want to show that points $z$ satisfying $|z - a| = k|z - b|$ (where $a$ and $b$ are also complex) gives me a circle of radius $$\frac{k|a - b|}{|1-k^2|},$$ centred at $$\frac{a-bk^2}{1-k^2}.$$

I tried to boil this down to a cartesian equation that showed me the properties of the circle I was looking for, found the centre by doing this, but still (as you can imagine) ended up in a horrible algebraic mess, with no hope of showing my right hand side was in fact the square of the radius.

Is there a "neater" way of deriving that we have a circle with the above centre and radius?

Many thanks! Sam

  • $\begingroup$ Use $\LaTeX\text{}$ $\endgroup$
    – user93957
    Commented Nov 26, 2013 at 19:02

1 Answer 1


Presumably $k > 0$. To reduce the algebraic mess, take advantage of symmetry. You want the centre to be at $a - k^2 b$, so let's translate so that is $0$.
Then rotate and scale so that $b=1$ and $a = k^2$. Now your equation is $|z - k^2| = k |z - 1|$. Squaring both sides, $$ (z - k^2)(\overline{z} - k^2) = k^2 (z - 1)(\overline{z} - 1)$$ Now expand and simplify.

  • $\begingroup$ Thanks. So I get this down to a circle, centre 0 and radius k|1-k^2|/|1-k^2| which is good. But how do I do the scaling and rotation to get to this point? $\endgroup$
    – user111665
    Commented Nov 26, 2013 at 19:36
  • $\begingroup$ Divide by $b$. That is, $|z - a| = k |z - b|$ iff $|w - a/b| = k |w - 1|$, where $w = z/b$. Note that this mapping preserves circles. $\endgroup$ Commented Nov 26, 2013 at 21:43

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