Proving that the lines cut the coordinate axes in concylic points The lines $2x +3y +19 = 0$ and $9x+6y-17 = 0$ cut the coordinate axes in concyclic points.What would be the fastest method to prove it manually?Is it possible to prove the statement without having to find the equation of the circle?
 A: 
$A,B,C,D$ lie on the same circle iff $$OA\cdot OB = OC\cdot OD,$$
i.e. iff $OAD$ and $OCB$ are similar triangles. 
The lengths $OA,OB,OC,OD$ are straightforward to compute by setting $x=0$ or $y=0$ in the equations of the given lines. That ultimately boils down to: 

Since $\color{red}{3\cdot 6=2\cdot 9}$, the lines $2x+3y+19=0$ and $9x+6y-17=0$ meet the coordinates axis at concyclic points.

A: If you use Ptolemy's theorem, you can prove it by find intersection points with axist and compute some length with Phytagorean theorem. (You don't have to find center of circle.)
A: Let the two given line be $L_1$ and $L_2$.

From $L_2$, find $\alpha$ and hence $\alpha ‘$ is known.
Similarly, find $\phi$.
Is $\alpha ‘ = \phi$?
A: Your four points are 
$\left(0\mid-\frac{19}3\right);
\left(-\frac{19}2\mid 0\right);
\left(0\mid\frac{17}6\right);
\left(\frac{17}9\mid 0\right);
$    
If 
$\begin{vmatrix}
0^2+\left(-\frac{19}3\right)^2&0&-\frac{19}3&1\\
\left(-\frac{19}2\right)^2+0^2&-\frac{19}2&0&1\\
0^2+\left(\frac{17}9\right)^2&0&\frac{17}9&1\\
\left(\frac{17}9\right)^2+0^2&\frac{17}9&0&1\\
\end{vmatrix}=0$
then your four points are concyclic.    
Since the RHS is 0, the above determinant can be simplified into
$\begin{vmatrix}
19^2&0&-19·3&3^2\\
19^2&-19·2&0&2^2\\
17^2&0&17·6&6^2\\
17^2&17·9&0&9^2
\end{vmatrix}$ 
