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Suppose I have two algebraic curves $C_1$ and $C_2$ in the plane.


DistAlgCurves
I would like to find the minimum distance between the two curves. If the two curves have degrees $n_1$ and $n_2$, what is (generically) the degree of the equation one of whose roots identifies that minimum distance?

Above, $n_1=3, n_2=2$: $$ C_1 \;:\; y=x^3 $$ $$ C_2 \;:\; (x+\tfrac{5}{4})^2 + (y-\tfrac{3}{4})^2 = \tfrac{1}{2} $$

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  • $\begingroup$ How do you define degree here? $\endgroup$ – Kaster Jan 9 '15 at 2:20
  • $\begingroup$ Sounds like something that Bernd Sturmfels can give a pertinent answer to. $\endgroup$ – orangeskid Jan 9 '15 at 2:34
  • $\begingroup$ @Kaster: I meant the highest combined exponent of $x$ times $y$, e.g., $x^2 y^3$ is degree $5$. $\endgroup$ – Joseph O'Rourke Jan 9 '15 at 11:43
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I am not sure I answer exactly the question as posted; so, forgive me if what I write is out of topic.

If you have two curves defined by $y_1(x)$ and $y_2(x)$, the square of the distance between the two curves is given $$\Phi(x_1,x_2)=\big(x_1-x_2\big)^2+\big(y_1(x_1)-y_2(x_2)\big)^2$$ and you want to minimize this function with respect to $x_1$ and $x_2$.

The problem can be difficult with multi-valued functions (as in the case the the circle you gave). This can lead to very complex equations and, more than likely, only numerical methods will be able to do the job.

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