# Degree of min distance function between two algebraic curves

Suppose I have two algebraic curves $C_1$ and $C_2$ in the plane. 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, $n_1=3, n_2=2$: $$C_1 \;:\; y=x^3$$ $$C_2 \;:\; (x+\tfrac{5}{4})^2 + (y-\tfrac{3}{4})^2 = \tfrac{1}{2}$$

• How do you define degree here? – Kaster Jan 9 '15 at 2:20
• Sounds like something that Bernd Sturmfels can give a pertinent answer to. – orangeskid Jan 9 '15 at 2:34
• @Kaster: I meant the highest combined exponent of $x$ times $y$, e.g., $x^2 y^3$ is degree $5$. – Joseph O'Rourke Jan 9 '15 at 11:43

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 $x_2$.