# 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

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?

• 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? – Semiclassical Jun 11 '16 at 5:51
• Even I don't know about that. Actually I got the question from a practice book. – user335710 Jun 11 '16 at 5:54
• @user335710 desmos.com/calculator/5j4vz6bm8j – A---B Jun 11 '16 at 6:34
• 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. – Oscar Lanzi Jun 11 '16 at 10:30

## 2 Answers

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 $O$. • Could you please explain me abit more? It's not clear to me. – user335710 Jun 11 '16 at 8:25
• Ehh. I understood............................? – user335710 Jun 11 '16 at 9:11
• I added a picture in my answer, hoping that helps. – Ng Chung Tak Jun 11 '16 at 9:38
• 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? – user335710 Jun 11 '16 at 9:41
• 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 $\Delta=a^2 c - a f^2 - a g^2 - c h^2 + 2f g h\,$. – Ng Chung Tak Jun 11 '16 at 9:45

Begin by finding intercepts:

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

For x intercepts,

$$ay^2 + 2fy +c = 0$$

For y intercepts,

$$ax^2 + 2gx + c = 0$$

Let the points corresponding to root be: $$(a,0) , (b,0) , (0,c) , (0,d)$$

From this answer here, the equation of circle through these points:

$$\det\begin{bmatrix} x^2 +y^2 & x & y & 1 \\ b^2 & b & 0 & 1 \\ c^2 & 0 & c& 1 \\ d^2 & 0 & d& 1 \\ \end{bmatrix} = 0$$

For checking if $$(a,0)$$ is on this circle , simply plug it in place of $$(x,y)$$