TITLE: Question about concyclic points on the coordinate axes
QUESTION [8 upvotes]: If the points where the lines $3x-2y-12=0$ and $x+ky+3=0$ intersect both the coordinate axes are concyclic,then the number of possible real values of k is
(A)1 $\hspace{2cm}$(B)2$\hspace{2cm}$(C)3$\hspace{2cm}$(D)4
My attempt:I used Ptolemy theorem$AC\times BD=AB\times CD+BC\times AD$ but the calculations and simplifications are very messy and clumsy.Is there an elegant way of solving this problem?

REPLY [6 votes]: Lets take two Straight lines $$a_1x+b_1y=1$$ and $$a_2x+b_2y=1$$ which intersect Coordinate axes at four points which are Concyclic.Let the Circle that passes through these four points be $$x^2+y^2+2gx+2fy+c=0$$. Now since it cuts X axis at $\left(\frac{1}{a_1},0\right)$ and $\left(\frac{1}{a_2},0\right)$ , it is evident that $\frac{1}{a_1}$ and $\frac{1}{a_2}$ are roots of $$x^2+2gx+c=0$$, So product of the roots is $$\frac{1}{a_1a_2}=c$$. Similarly $\frac{1}{b_1}$ and $\frac{1}{b_2}$ are roots of $$y^2+2fy+c=0$$ with product of the roots as again
$$\frac{1}{b_1b_2}=c$$
Hence if  two Straight lines $$a_1x+b_1y=1$$ and $$a_2x+b_2y=1$$  intersect Coordinate axes at four points which are Concyclic, Then 
$$a_1a_2=b_1b_2$$