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Show that the four asymptotes of the curve $(x^2-y^2)(y^2-4x^2)+6x^3-5x^2y-3x^2y+2y^3-x^2+3xy-1=0$ cut the curve again in eight points which lie on a circle of radius unity.

I have found the asymptotes to be $y=x; y=-x-1; y=2x; y=-2x-1$. Then I found the combined equation of the 4 asymptotes : $(x^2-y^2)(y^2-4x^2)-3x^2y+xy^2+4x^3+y^2-3xy=0$

To find points of intersection I subtracted combined equation of asymptotes from equation of curve, but it does not give equation of unit circle. What am I missing? Please help.

user464068
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To find the intersections of the asymptotes and the curve you need to solve the equations simultaneously. Those will give the common points. Taking your first asymptote $y=x$ the first product is zero and we need to solve $$6x^3-5x^2y-3x^2y+2y^3-x^2+3xy-1=0\\$$ All the third power terms cancel and we are left with $$2x^2-1=0\\x=\frac {\sqrt 2}2=y$$ That is one point of intersection. Go through your asymptotes. Each one will reduce the curve equation to an equation in one variable. Solve each one. You should find a total of eight solutions.

Ross Millikan
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