I am facing trouble with the following question

What is the condition that the two conic sections $ax_1^2+2h_1xy+b_1y^2+2g_1x+2f_1y+c_1=0$ and $ax_2^2+2h_2xy+b_2y^2+2g_2x+2f_2y+c_2=0$ intersects each other in four concyclic points.

I took the four points as unknown terms and since the points satisfies both the equations,I tried to put the points in each of the equations and tried to solve for the coordinates but it became useless and I got nothing.

Is there some way to arrive at the condition for the concyclicity of the intersecting points?Any help would be appreciated.Thanks.


Eliminate the cross term:

$$a(h_2-h_1)x^2+(b_1 h_2-b_2 h_1)y^2+2(g_1h_2-g_2h_1)x+2(f_1h_2-f_2h_1)y+(c_1h_2-c_2h_1)=0$$

Equating $x^2$ and $y^2$ terms: \begin{array}{rclll} a(h_2-h_1) &=& b_1 h_2-b_2 h_1 && \\ h_2 &=& \dfrac{h_1(a-b_2)}{a-b_1} &,& (a\ne b_1) \\ b_2 &=& a &,& (a= b_1) \\ \end{array}

  • $\begingroup$ I didnot get how you arrived at the equation of the circle.Can you please elaborate it.Thanks. $\endgroup$ – Navin Dec 1 '16 at 14:08
  • $\begingroup$ The quadratic terms in circles must be in the form of $A(x^2+y^2)$. That's no cross terms $xy$ and same coefficients in $x^2$ and $y^2$. By the way, the two conics satisfy the condition not necessarily intersecting at four points. $\endgroup$ – Ng Chung Tak Dec 1 '16 at 14:09
  • $\begingroup$ Yes,I know that coefficient of $x^2$ and $y^2$ are the same.but how did you arrived at the equation of the circle passing through the four points of intersection. $\endgroup$ – Navin Dec 1 '16 at 14:11
  • 1
    $\begingroup$ For intersection, $(x,y)$ satisfies both $C_1=0$ and $C_2=0$ and hence their linear combination $\lambda C_1+\mu C_2=0$ (which is a family of conics also passing through those intersections). If there exists $\lambda$, $\mu$ such that $\lambda C_1+\mu C_2=0$ is a circle, then the points are concyclic. $\endgroup$ – Ng Chung Tak Dec 1 '16 at 14:15

Given $4$ points, the family of conics $ax^2+hxy+by^2+gx+fy+c = 0$ going through them is given by a projective line (it gives $4$ linear equations on the coefficients of the conic).

(and conversely, given $2$ conics in generic position, they meet in $4$ points, and the projective line joining those two conics give you equations of conics that also go through those points, so every projective line of conics correspond to $4$ points)

And such a conic is a circle when $h=0$ and $a=b$.

Thus, if you have two conics, they intersect at $4$ cocyclic points if and only if there is an equation of a circle in the projective line joining the two equations.

This means there exists $s,t$ not both zero such that $sh_1+th_2 = 0$ and $sa_1+ta_2 = sb_1+tb_2$.

If for example $h_1$ or $h_2$ is nonzero we have to pick $t = -h_1$ and $s=h_2$ (up to a multiplicative constant), and so we need $h_2a_1-h_1a_2 = h_2b_1-h_1b_2$


Please see whether this link helps you.


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