I was plotting some equations and I got with the curious relation

If we build the triange enter image description here Such that it follows the following relation: $$AD=a$$ $$DB=b$$ $$AC=ak$$ $$CB=bk$$

Then when we vary $k$ from the smallest possible value to the greatest possible one according to triangle inequality, the point C plots a circle around the midpoint of the maximum and minimum solutions, with it's diameter being the segment between them.

In a more formal way

If $C$ is the center of the hypotetic circle, $$\frac{AD}{DB}=\frac{AE}{EB}$$ and according to triangle inequality $$\frac{AD}{DB}=\frac{FA}{FB}$$ Prove that $$CD=CE$$ That is, prove that the point E in fact lies in the hypotetic circle. enter image description here This came out of curiosity when i plotted the point C as k varies, I was expecting a more weird shape but I got with a circle!.I have empirically verified it in geogebra and such, but i have not been able to prove it.The closest i have gotten is that $CA=\frac{AD^2}{DB-AD}$, imposing that $DB>AD$. If they are equal, a division by cero occurs, but if $DB<AD$, the circles moves to the other side.


Here's a brute-force coordinate approach ...

Edit. My point names aren't consistent with OP's (which vary between his diagrams anyway :). In case of edits to the question, I'll explain the roles of my points here.

We take point $D$ on segment $AB$, with $a := |AD|$ and $b := |BD|$. A variable point $P$ is defined by $|AP|=k |AD|$ and $|BP|=k|BD|$. Note that, if $a=b$ the locus of $P$ is the perpendicular bisector of $AB$ (which happens to be the limiting case of circles), so we take $a > b > 0$.

Coordinatizing with points $A( -a, 0 )$, $B( b, 0 )$, $D( 0, 0 )$, $P(x,y)$, we have $$\begin{align} |AP| &= k |AD|: \qquad ( x + a )^2 + y^2 = a^2 k^2 \\ |BP| &= k |BD|: \qquad ( x - b )^2 + y^2 = b^2 k^2 \end{align}$$

Equating $k^2$ obtained from both equations gives $$\frac{(x+a)^2 + y^2}{a^2}=k^2=\frac{(x-b)^2+y^2}{b^2}$$ so that $$b^2 (x+a)^2 + b^2 y^2 = a^2 ( x-b)^2 + a^2 y^2$$

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

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

Since the (non-zero!) coefficients of $x^2$ and $y^2$ match (and there's no $xy$ term), we definitely have a circle; without a constant term, we see that the circle passes through the origin (point $D$). For more information, we put the equation into standard form:

$$\left( x - \frac{a b}{a-b} \right)^2 + y^2 = \left(\frac{ab}{a-b}\right)^2$$

So, the circle has center $C = \left(\frac{ab}{a-b},0\right)$ and radius $|CD| = \frac{ab}{a-b}$, and we can verify your conjecture:

$$|CA| = \frac{ab}{a-b}+a = \frac{a^2}{a-b} \qquad |CB| = \frac{ab}{a-b}-b=\frac{b^2}{a-b}$$

  • $\begingroup$ Thank you! I was very curious about it. Also, this "theorem" helps me a lot to simplify my equations for the app I was developing and plotting around. $\endgroup$ – chubakueno May 31 '13 at 1:07

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