Question on geometric inversions Consider triangle ABC such that AB = 3 ,BC = 4 ,CA = 5. Find the
point O such that after an inversion centered at O the line passing through A,C becomes a
line, and lines passing through A,B and through B,C become circles of equal radii.
I've been trying to work on this problem and from my knowledge I know that the center of inversion must be somewhere on the line AC since its transforming into a line. i also know that this center O of inversion must be equidistant somehow from the lines AB and BC but Im not sure how exactly to find the point geometrically that the same distance from both of those points? Any hints or help would be greatly appreciated
 A: In fact, your analysis is good ; the only point that has been lacking in it is that a triangle with such lengthes $3-4-5$ is a right triangle in $B$. We can take profit of this orthogonality by  installing the triangle in the natural system of coordinates that can be seen in the following picture.
In this way, taking into account your remarks, the inversion center $\Omega$

*

*must belong to line $BC$ and


*must be at equal distance from lines $AB$ and lines $BC$, which means that $\Omega$ belongs to the angle bissector in $B$.
Therefore $\Omega$ is situated at the intersection of these two straight lines i.e., its coordinates are the solution to the system:
$$\begin{cases}y&=&-x\\y&=&\frac43x+4\end{cases}$$
giving
$$\Omega \left(-\frac{12}{7},\frac{12}{7}\right)$$

A: $O$ lies on $AC$ and also the perpendicular $OC_1$ dropped from $O$ to $AB$ and the perpendicular $OA_1$ dropped from $O$ to $BC$ are equal. By, similarity of $\triangle AC_1O$ and $\triangle OA_1C$, $\frac{AC_1}{OA_1}=\frac{C_1O}{A_1C}$ which gives $\frac{3-r}{r}=\frac{r}{4-r}$. Thus $r=\frac{12}{7}$ and $O$ is the unique point on $AC$ which satisfies $\frac{AO}{OC}=\frac{3}{4}.$
