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Old May 19th, 2008, 10:15 AM
potticus potticus is offline
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Default intersection of a circle and a cubic equation

Other people have given solutions for a circle and a parabola, but they mainly involve finding a quadratic in x^2, suitable as they want 2 solutions.

In this case, I need a single solution to a circle/cubic problem. I know there's only one solution because the radius of the circle has been set to the smallest distance between the line and the centre point.

I tried substituting the cubic equation into the equation for the circle, then after a lot of bracket expanding (which I've checked over and over), I still don't get any intersection points.

I end up with a 6th order equation (which I believe to be correct - just thinking logically, it seems there can be 6 intercepts between a cubic and a circle). However, finding the roots of this equation (using the roots function in matlab) doesn't work :s

This is where i'm at so far:

With (x-xp)^2 + (y-yp)^2 = dc^2
And y = ax^3 + bx^2 + cx + d

After substitution, expansion and simplification:
0=(a^2)x^6 + (2ab)x^5 + (2ac + b^2)x^4 + (2ad + 2bc - 2ayp)x^3 + (2bd + c^2 - 2byp + 1)x^2 + (2cd - 2cyp -2xp)x + (d^2 + yp^2 - 2dyp + xp^2 - dc^2)

I can't see why this doesn't give me a suitable value for x intercept point. Can someone please help me out?

Thanks
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