Let $A=\{(x,y)∈R^2: x^2+y^2\le4\}$ and $B=\{(0,y)∈R^2:|y|\le3\}$. Determine Hausdorff distance between $A$ and $B$.
I wrote $d(2)(B,A)=((0,3),A)=((0,3)(0,2))=1$. What about $d(A,B)$? $d(2)(A,B)=((2,0),B)=((2,0)(0,0))=2$. Am I right here?
Then the Hausdorff distance would be $2$
Your answer is correct but your solution needs some clarification.
Since $A$ and $B$ are closed you need to find the point $a \in A$ that is furthest from $B$ and the point $b \in B$ that is furthest from $A$. You can draw a picture to see that $\max \mathrm{dist}(a,B) = 2$ which is attained at $(2,0)$ and $(-2,0)$, and that $\max \mathrm{dist} (b,A) = 1$ which is attained at $(0,3)$ and $(0,-3)$. The Hausdorff distance is $$d(A,B) = \max \left\{ \max \mathrm{dist}(a,B) , \max \mathrm{dist} (b,A) \right\} = 2.$$