Here is a circle with radius 3 as you may have expected. On the same
set of axes , graph
Now we have an ellipse on top of the hyperbola. Notice that there are
four intersection points, all 3 units from the center of the circle.
Let's try changing the coefficient of the xy term to 2.
What a surprise! The lines appear to be parallel, but notice that they
pass through the same four intersections that the circle and the ellipse
do. Let's try a coefficient of 4 for the xy term.
Here's another surprise: a hyperbola. Let's try some larger numbers,
say 8 and 12 .
That just gives us sharper hyperbolas. What will happen if we try a number
between l and 2 ? Let's try 3/2.
Now we have a more elongated ellipse. What will happen if we try one
number closer to 1, say 9/8 and one closer to 2, say 19/10.
So the 9/8 xy is close to the first ellipse with coefficient 1. And the 19/10 gives a very long ellipse which appears to be approaching the parallel lines when the coefficient is 2. It seems that 2 sets a boundary. What will happen if we try some negative coefficients? Try -1,-2, and -6 for coefficients.
The graphs are rotated 90 degrees clockwise. But what will happen if we change the radius of the circle? Will the boundary be different? Let's try the equations
The pattern is the same, and the boundaries are still 2 and -2. The conjecture is that the radius of the circle does not matter, that all xy terms with coefficient of 2 will produce parallel lines through the circle at four intersection points.