Distance between a point $x \in \mathbb R^2$ and $x_1^2+x_2^2 \le 4$

Question

Let $S=\{(x_1,x_2):x_1^2+x_2^2\le 4\}$. Let $f(y)=\inf\{||y-x||:x\in S\}$. Find $f$ explicitly.

This is the $f(y)$ I got so far

$$\inf \left\{ \sqrt{(y_1-x_1)^2+(y_2-x_2)^2}:(x_1+x_2) \in S \right\}$$

How do I relate $f(y)$ with $x_1^2+x_2^2\le 4\}$?

Answer 1

$$f (\mathrm y) := \max \left( 0, \|\mathrm y\|_2 - 2 \right) = \begin{cases} 0 & \text{if } \|\mathrm y\|_2 \leq 2\\ \|\mathrm y\|_2 - 2 & \text{if } \|\mathrm y\|_2 \geq 2\end{cases}$$

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Proof

The case $\|\mathrm y\|_2 \leq 2$ is trivial. If $\|\mathrm y\|_2 \geq 2$, the squared Euclidean distance between $\rm y$ and the disk is equal to the squared Euclidean distance between $\rm y$ and the disk's boundary, i.e., a circle of radius $2$.

$$\begin{array}{ll} \text{minimize} & \| \mathrm y - \mathrm x \|_2^2\\ \text{subject to} & \| \mathrm x \|_2 = 2\end{array}$$

Note that

$$\begin{array}{rl} \| \mathrm y - \mathrm x \|_2^2 &= \| \mathrm y \|_2^2 - 2 \langle \mathrm x , \mathrm y \rangle + \| \mathrm x \|_2^2 \\ &\geq \| \mathrm y \|_2^2 - 2 \| \mathrm x \|_2 \| \mathrm y \|_2 + 4\\ &= \| \mathrm y \|_2^2 - 4 \| \mathrm y \|_2 + 4\\ &= \left( \| \mathrm y \|_2 - 2 \right)^2\end{array}$$

where the Cauchy-Schwarz inequality was used. Hence, $$\| \mathrm y - \mathrm x \|_2 \geq \| \mathrm y \|_2 - 2$$

Answer 2

Basic approach. Do this geometrically. Visualize $S$ as a set on the Cartesian plane with coordinates $(x_1, x_2)$. What does this set look like? Then observe that $f(y)$, for any point $y$ in the plane, denotes the infimum (the greatest lower bound, in other words) of the distance from $y$ to any point in $S$. That is to say, $f(y)$ is the length of the shortest line joining $y$ to $S$.

Can you give an explicit expression for the length of this line?

Answer 3

The answer is $f(y)= 0 $ if $y\in S$ obviously. If $y\notin S,$ due to the symmetry of $S$ we have that $f(y)=(1-\lambda)\|y\|,$ were $\lambda$ is the greatest positive number such that $\lambda y \in S.$ We have $\lambda y \in S$ iff $\|\lambda y\|\leq 2,$ whcih results in $\lambda = \frac{2}{\|y\|}.$ Hence, you have

$$f(y)= (1-\frac{2}{\|y\|})\|y\|= \|y\|-2,$$ if $y\notin S.$