Showing metric is a square

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Consider $\mathbb{R}^{2}$.

One metric on this set is $d\left ( x,y \right )=\max\left \{ \left | x_{1}-y_{1} \right |,\left | x_{2}-y_{2} \right | \right \}$, for which the open balls $B_{\epsilon}\left ( a \right )$ are squares with sides of length $2\epsilon$, parallel to the coordinate axes.

Another metric is $d\left ( x,y \right )=\max\sqrt{\left ( x_{1}-y_{1}\right )^{2}+\left ( x_{2}-y_{2} \right )^{2}}$, for which the open balls are circles of radius $\epsilon$ centred at a.

Could someone show me why the first is a square of sides 2$\epsilon$ and the second is a ball of radius $\epsilon$? Hint for the second.

Thanks in advance.