Intersection of a hyperbola and a circle

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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;
 Manipulate[
    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}]

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}}