The area of the region bounded by the locus of a point $$P$$ satisfying $$d\left( {P,\,A} \right) = 4,$$   where $$A$$ is $$\left( {1,\,2} \right)$$  is :

Solution :
We have, $${\text{max}}\left\{ {\left| {x - 1} \right|,\,\left| {y - 2} \right|} \right\} = 4$$
If $$\left\{ {\left| {x - 1} \right| \geqslant \left| {y - 2} \right|} \right\},$$
then $$\left| {x - 1} \right| = 4,$$
i.e., if $$\left( {x + y - 3} \right)\left( {x - y + 1} \right) \geqslant 0,$$
then $$x = - 3{\text{ or }}5$$
If $$\left| {y - 2} \right| \geqslant \left| {x - 1} \right|,$$
then $$\left| {y - 2} \right| = 4$$
i.e., $$\left( {x + y - 3} \right)\left( {x - y + 1} \right) \leqslant 0,$$
then $$y = - 2{\text{ or }}6$$
So, the locus of $$P$$ bounds a square, the equation of whose sides are $$x = - 3,\,x = 5,\,y = - 2,\,y = 6$$
Thus, the area is $${\left( 8 \right)^2} = 64.$$