Geometrical place of one point

Question

Through one of the intersection points of two given circles, a line is build which intersects 2nd time the 2 circles in $A$ and $B$. Determine the geometrical place of the middle of $AB$.

thanks for your time and help!

Answer 1

We know that $\angle BDG$ and $\angle BEG$ are constant, because $BG$ is constant. Therefore, all angles in $\triangle GDE$ are constant. Thus, $\triangle GDE\sim \triangle GD'E'$ for any other choice of $DE$. There is a linear transformation $T$ of the plane which sends $D$ to $F$. This is a rotation around $G$ with some constant angle $\angle DGF$. It also scales the plane with a constant factor $\frac{|FG|}{|DG|}$. This is a linear transformation. We can 'generate' all possible lines $DE$ by letting $D$ move over the circle with center $A$. Now, we now that $F$ is the image of $D$ after applying $T$ to it. Thus, the image of $D$ when it moves over the circle is the image of that circle when applying $T$ to it. Because $T$ is a linear transformation, it preserves shape, so the image will be a circle through $G$ again. We can also take the image of $A$. It is not hard to see that we can choose $DE$ in such a way that $A$, $D$, and $G$ and $E$, $C$ and $G$ are collinear. We know that $F$ is the center of $DE$, and because $|AB|=\frac 12|DG|$ and $|CG|=\frac12|GE|$, we know that $H$ will be the center of $AC$. Thus, the image of the circle with center $A$ after $T$ is the circle through $B$ and $G$ with center the midpoint of $AC$.