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Show that pair of straight lines $ax^{2}+2hxy+ay^{2}+2gx+2fy+c=0$ meet coordinate axes in concyclic points. Also find equation of the circle through those cyclic points

My Attempt:

Given equation to the pair of straight lines is

$ax^2+2hxy+ay^2+2gx+2fy+c=0$

Let the lines be

$l_1x+m_1y+n_1=0$ and $l_2x+m_2y+n_2=0$

Now what should I do next?

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  • $\begingroup$ The equation in your title is quadratic in $x$ and $y$, and so is a conic section not a pair of straight lines. Do you mean for the lines to be tangents to this conic? $\endgroup$ – Semiclassical Jun 11 '16 at 5:51
  • $\begingroup$ Even I don't know about that. Actually I got the question from a practice book. $\endgroup$ – user335710 Jun 11 '16 at 5:54
  • $\begingroup$ @user335710 desmos.com/calculator/5j4vz6bm8j $\endgroup$ – A---B Jun 11 '16 at 6:34
  • $\begingroup$ First off, for the coefficients of $x^2$ and $y^2$ to be equal you have to have $l_1l_2=m_1m_2$. A completely general pair of lines does not intersect the axes at concyclic points. $\endgroup$ – Oscar Lanzi Jun 11 '16 at 10:30
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Put $x=0$, $$ay^2+2fy+c=0$$

$$y_{1}y_{2} = \frac{c}{a}$$

Put $y=0$, $$ax^2+2gx+c=0$$

$$x_{1}x_{2} = \frac{c}{a}$$

Hence, $$x_{1}x_{2} = y_{1}y_{2}$$

By Converse of Intersecting chord theorem, the intercepts are concyclic.

Note briefly:

Note that it's also true for any non-degenerate conics such that $g^2>ac$, $f^2>ac$ and $c\neq 0$.

For two straight lines,

$$\begin{vmatrix} a & h & g \\ h & a & f \\ g & f & c \end{vmatrix}=0$$

Now the equation of the required circle is

$$\fbox{$a(x^2+y^2)+2(fx+gy)+c=0$}$$

See the link here.

In the diagram below, $A=(x_{1},0), B=(x_{2},0), C=(0,y_{1}), D=(0,y_{2})$, $AB$ and $CD$ are the intersecting chords that meet at the origin $O$.

enter image description here

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  • $\begingroup$ Could you please explain me abit more? It's not clear to me. $\endgroup$ – user335710 Jun 11 '16 at 8:25
  • $\begingroup$ Ehh. I understood............................? $\endgroup$ – user335710 Jun 11 '16 at 9:11
  • $\begingroup$ I added a picture in my answer, hoping that helps. $\endgroup$ – Ng Chung Tak Jun 11 '16 at 9:38
  • $\begingroup$ Yes. It a bit more clear. But I am not familiar with finding the determinant of $3\times 3$ matrix. Could you make me understand that? $\endgroup$ – user335710 Jun 11 '16 at 9:41
  • $\begingroup$ In general for conic $a x^2+2h xy+b y^2+2gx+2fy+c=0\,$, $\Delta=\begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix}=a b c - a f^2 - b g^2 - c h^2 + 2 f g h\,$. In this case $\Delta=a^2 c - a f^2 - a g^2 - c h^2 + 2f g h\,$. $\endgroup$ – Ng Chung Tak Jun 11 '16 at 9:45

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