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If the points where $ 3x-2y-12=0$ and $x+ky+3=0$ intersect both the co-ordinate axes are concyclic. Then the number of possible real values of $k$ are?

The first equation cuts the axes at $(4,0)$ and $(0,-6)$. While the second equation cuts the coordinate axes at $(-3,0)$ and $(0,-\frac{3}k)$. Since the first 3 points I mentioned are fixed. Only one circle will pass through them. This one circle will cut the $y$ axis at only one point.

Hence, I expect only one value of $k$ for the condition above to happen. However, the answer is that four values of $k$ are possible.

Please do point out the mistake in my solution.
Any help will be appreciated.

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  • $\begingroup$ The formulation is hard to understand. Perhaps we first have to intersect the lines , and then we search a circle touching both axes and going through the intersection-point. I admit, it is a guess. $\endgroup$
    – Peter
    Commented Apr 2, 2018 at 13:03
  • $\begingroup$ A longer formulation with two sentences would probably be much clearer. $\endgroup$
    – Peter
    Commented Apr 2, 2018 at 13:08
  • $\begingroup$ @Peter I'm open for suggestive edits. $\endgroup$
    – SmarthBansal
    Commented Apr 2, 2018 at 13:16
  • $\begingroup$ Unfortunately, I am not sure whether I guessed the intented question right. $\endgroup$
    – Peter
    Commented Apr 2, 2018 at 13:17
  • $\begingroup$ What do you mean by cyclic? Or you meant concyclic? $\endgroup$
    – Rohan Shinde
    Commented Apr 2, 2018 at 15:19

2 Answers 2

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Four values for $K$ is implausible, but there are certainly two possible values for $K$. The circle through the three fixed intercepts crosses the $y$-axis at two places, not one, because the segment from $(-3,0)$ to $(4,0)$ is a chord of the circle. We already know one of these intercepts: it’s the $y$-intercept of the fixed line. I suppose that solution could be rejected on the grounds that the four points must be distinct. I’m curious to see the purported solution. Perhaps we’re all misinterpreting the problem.

The points are concyclic if $$\det\begin{bmatrix}0^2+\left(-\frac3K\right)^2 & 0 & -\frac3K & 1 \\ 4^2+0^2 & 4 & 0 & 1 \\ 0^2+(-6)^2 & 0 & -6 & 1 \\ (-3)^2+0^2 & -3 & 0 & 1\end{bmatrix}=0.$$ Expanding and solving for $K$ yields $K=\frac12$ or $K=-\frac32$. For $K=\frac12$, the $y$-intercept is $-6$, our already-known point, so the only new one is $(0,2)$ for $K=-\frac32$.

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I think there is only one value of $K$, and this is how I found it. If you could tell us the values of $K$ in the answer, it would be helpful.

If $X_1, X_2$ and $Y_1,Y_2$ are the intersection points of line 1,2 in the $X$ and $Y$ axis respectively, then by the power of the origin with respect to the circle passing through $X_1, X_2,Y_1,Y_2$ we must have $$OX_1OX_2=OY_1OY_2$$ Which gives $$-12=\frac{18}{K}$$ $$K=\frac{-3}{2}$$

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  • $\begingroup$ I got the same answer! I don't see where the other 3 come from. $\endgroup$
    – SmarthBansal
    Commented Apr 2, 2018 at 13:33

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