Suppose that we have algebraic equations for a circle and a hyperbola given by $$x^2 + y^2 = 1$$ $$Ax^2 + 2Bxy - Ay^2 + Cx + Dy = 0$$ respectively. The real coefficients $A,B,C,D$ have a negative discriminant $\Delta = -(A^2 + B^2)$. These two curves intersect due to the origin being a root to the hyperbola equation, which can be viewed as an unbounded function in the $xy$-plane.

Does there exist algebraic geometric methods for finding the intersection points? I have little knowledge of the subject but if there are methods for specifying the intersection points of these simple curves I would like to know. I know that substitution leads to a fourth order polynomial but solution by radicals leads to complicated formula which seem intractable. I am interested in finding these intersection points as they correspond to equilibrium points of a system of differential equations. I am curious if anyone knows a change of coordinates that make finding these intersections a little more tractable.

  • $\begingroup$ You can find a general recipe for intersecting two conics here. It entails finding a degenerate linear combination of them (which involves solving a cubic), splitting them into their component lines, and then intersecting those lines with one of the original conics. $\endgroup$ – amd Mar 29 '19 at 17:52
  • $\begingroup$ For this case in particular, you could instead find a parameterization of the hyperbola and then solve for values of the parameter for which the distance to the origin is $1$. $\endgroup$ – amd Mar 29 '19 at 17:52
  • $\begingroup$ @TheRefrigerator Is there an answer you like here? If you do, please accept it. $\endgroup$ – Quote Dave Jun 18 '19 at 16:49

Going off of @Peter Foreman's answer, you can rewrite the second equation as $$\dfrac{2bx+d\pm\sqrt{4b^2x^2+4bdx+d^2+4a^2x^2+4acx}}{2a}=y$$. Then, rewrite the first equation as $$y=\pm\sqrt{1-x^2}$$ Make the two equations equal to each other and solve.


Write $y=\pm\sqrt{1-x^2}$ and use this definition in the second equation. This problem then becomes solving a quartic for which there is a quartic formula.

  • $\begingroup$ It is $$y=\pm\sqrt{1-x^2}$$ $\endgroup$ – Dr. Sonnhard Graubner Mar 29 '19 at 16:37
  • $\begingroup$ Yes, sorry it is. $\endgroup$ – Peter Foreman Mar 29 '19 at 16:38

May not be the answer you are looking for, and probably not the forum for it, but you can plot the two curves using software, for example Mathematica or MATLAB or Python or Java or ...

Here is a quick way to do it in Mathematica, for example:

 a[x_, y_] := (x^2 + y^2 - 1);
 b[x_, y_][A_, B_, C_, D_] := (A x^2 + 2 B x y - A y^2 + C x + D y );
 c = 3;
    Show[ContourPlot[{a[x, y], b[x, y][A, B, C, D]}, {x, -c, c}, {y, -c, c}],
      Graphics[{Red, PointSize -> Large, Point[{0, 0}]}]], 
        {A, -2, 2}, {B, -2, 2}, {C, -2, 2}, {D, -2, 2}]

enter image description here

You can also explicitly ask for the solutions:

With[{A = 1.15, B = -1.29, C = 1.95, D = 0}, 
  NSolve[{a[x, y] == 0, b[x, y][A, B, C, D] == 0}]]

This returns:

{{x -> 0.689595, y -> 0.724195}, {x -> 0.230435, y -> -0.973088}, 
 {x -> -0.871638, y -> 0.490151}, {x -> -0.799246, y -> 0.601004}} 

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