Compute the minimum distance between the centre to the curve $xy=4$. I wish to solve the following problem: 

Compute the minimum distance between the center to the curve $xy=4$.

But I don't know where to start from?
 A: We interpret the question as asking for the minimum distance from the origin to the curve $xy=4$. This is a rectangular hyperbola, with "centre" the origin.
The easiest way is to plot the curve, and note the symmetry. It is clear that the minimum is reached where $x=y$. Thus the nearest points to the origin are $(2,2)$ and $(-2,-2)$. The minimum distance is therefore $2\sqrt{2}$.
If we want to use calculus, note that the distance from $(x,y)$ to the origin is $\sqrt{x^2+y^2}$. We want to minimize $\sqrt{x^2+y^2}$ given that $xy=4$, that is, that $y=\frac{4}{x}$.
So we want to minimize $\sqrt{x^2+\frac{16}{x^2}}$, or equivalently to minimize its square. Let
$$f(x)=x^2+\frac{16}{x^2}.$$
We want to minimize $f(x)$.
Note that
$$f'(x)=2x-\frac{32}{x^3}.$$
Set this equal to $0$ and solve for $x$. We get $x=\pm 2$.
We need to verify that at $x=\pm 2$ we actually get a minimum. This can be shown in various ways. For example, for $x=2$, we can note that $f'(x)$ is negative in $(0,2)$, and positive in $(2,\infty)$. So in the interval $(0,2)$, the function $f(x)$ is decreasing, and thereafter it is increasing, so it must have reached a (local) minimum at $x=2$.
A: You want to minimise $D^2=x^2 + y^2$ subject to the condition $xy=4$.
$D^2=x^2 + \frac{16}{x^2}\ge2\sqrt{16}=8$ by AM-GM.
So the minimum distance is $\sqrt{8}$
A: You need to minimize the function
$$d= \sqrt{x^2 +y^2} = \sqrt{x^2+ 16/x^2}.$$
Which we got by considering the distance from a point on the curve to the origin. Now, use the derivative test. See related problem.
