Another way would be as following, though its incomplete at the end: For a triangle $\triangle{ABC}$ (assume $C\geq A \geq B$). Then the center of $k1$ is the intersection of the lines $l_{11}, l_{12}$ which passes bisects perpendicularly the lines $AO$ and $OA'$ respectively. As $A'$ is the reflection of $A$ about $N$, and $N$ is the middle of $HO$ ($H$ being the orthocenter), $AOA'H$ is a parallelogram, hence $OA'||AH$. So, $OA'$ is perpendicular to $BC$ and intersects $BC$ at the middle at $D$. Also we know, $AH=2OD$, which implies that $OD=DA'$. Hence, the center of $k1$ is the point where the perpendicular bisector of $AO$ cuts the line $BC$, let it be $\vec{C_1}=(x_1,y_1)$.

Let, $l_{12}$ upon extension intersects $AB$ at $T_1$. the, $\overline{AT_1}=\frac{R}{2}\csc{C}$ ($R$ is the circumradius of $\triangle{ABC}$). Also, $\angle{AT_{1}C_1}=C$. Then, similar calculation gives us the position vectors of $\vec{C_i}, i=1,2,3$, taking $B$ as the origin and $\vec{AB}$ line as $+x$ axis.

$\vec{C_1}=\left(\frac{(\frac{R}{2}\csc{C}-c)\cot{B}}{\cot{B}-\cot{C}},\frac{(\frac{R}{2}\csc{C}-c)}{\cot{B}-\cot{C}}\right)$

$\vec{C_2}=\left(-\frac{a\sin{A}-\frac{R}{2}}{\sin{(A-B)}} , 0\right)$

$\vec{C_3}=\left(-c-\frac{(\frac{R}{2}\csc{C}-c)\cot{A}}{\cot{A}-\cot{C}},\frac{(\frac{R}{2}\csc{C}-c)}{\cot{A}-\cot{C}}\right)$

As these circles have one common intersection point $O$, the other common would simply require that $C_1,C_2,C_3$ to be colinear. Hence, showing $\frac{y_3}{y_1}=\frac{x_3-x_2}{x_1-x_2}$ would prove it. $\frac{y_3}{y_1}=\frac{\cot{B}-\cot{C}}{\cot{A}-\cot{C}}$.