Question about concyclic points on the coordinate axes 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?
 A: 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 
$$a_1a_2=b_1b_2$$
A: The intercept-intercept forms of the lines are as follows:
$$\begin{align}
\frac{x}{4} + \frac{y}{-6} &= 1 \qquad\to\qquad
\cases{ x\text{-intercept: }\;\phantom{-}4 \\ y\text{-intercept: }\;-6}\\[6pt]
\frac{x}{-3} + \frac{y}{-3/k} &= 1 \qquad\to\qquad\cases{ x\text{-intercept: } \;-3 \\ y\text{-intercept: }\;-3/k }
\end{align}$$
So, we have this situation, with $\overleftrightarrow{AB}$ being the first line, and $C$ being a point on the second line:

Note that the question asks for the number of values of $k$ that solve the problem; that is, the number of points on the $y$-axis that are concyclic with $A$, $B$, $C$. Clearly,

Points $D$ and $B$ satisfy the description, so the answer is $2$.


While we don't actually need to determine the associated values of $k$, it's pretty easy to do.


*

*For solution point $B$, the $y$-intercept is $-6$; thus, $-3/k = -6$, so that $k = 1/2$.

*For solution point $D$, we calculate the power of the origin point with respect to the circle in two ways, and set them equal:
$$\begin{align}
|\overline{OA}|\;|\overline{O C}| &= |\overline{OB}|\;|\overline{OD}| \\
4\cdot 3 &= 6 \cdot |\overline{OD}| \\
2 &= |\overline{OD}|
\end{align}$$
Since $D$ is at distance $2$ from the origin, and it has a positive $y$-coordinate, we have $-3/k = 2$, so that $k = -3/2$.
A: hint that might be a bit easier
Let A and B be the $x-y$ intercepts of the line $3x-2y=12$ and C,D be those of the line $x+ky=-3$. Then using slopes compute the tangent of the angles ACB and ADB and use the fact that if these angles are equal then the points are concyclic.
A: From $L_{(1)}$, get $\alpha$.
$\beta$ is then known.
$\beta = \gamma$ (by angles in the same segment)
$\gamma$ is the angle of inclination of the required line.
Therefore $k$ can be found. 

Note that P, Q and R are fixed. Through 3 fixed points (on three different sections of the co-ordinate axes), only one circle can be drawn. If S (also on an axis) is the 4th con-cyclic points, other variants --- M (by varying k to get the red dotted line, example) or N (by the green dotted line), can never be the possible fourth. 
However, another possible candidate is when S is precisely Q, a degenerate case.

