Points $A$ and $B$ have coordinates $(2, 2)$ and $(6, 6)$. Find the coordinates of a point $P$ such that $PA = PB$ and the area of $\Delta PAB = 4$.

(A) Let the coordinates of point $P$ be $(x, y)$.
Given that $PA = PB$,we have $PA^2 = PB^2$.
Using the distance formula: $(x - 2)^2 + (y - 2)^2 = (x - 6)^2 + (y - 6)^2$.
Expanding both sides: $x^2 - 4x + 4 + y^2 - 4y + 4 = x^2 - 12x + 36 + y^2 - 12y + 36$.
Simplifying: $8x + 8y = 64$,which gives $x + y = 8$ (Equation $1$).
The area of $\Delta PAB$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 4$.
Substituting the coordinates $P(x, y)$,$A(2, 2)$,and $B(6, 6)$: $\frac{1}{2} |x(2 - 6) + 2(6 - y) + 6(y - 2)| = 4$.
$\frac{1}{2} |-4x + 12 - 2y + 6y - 12| = 4$.
$\frac{1}{2} |-4x + 4y| = 4$,which simplifies to $|-x + y| = 2$.
This implies $-x + y = 2$ or $-x + y = -2$ (Equation $2$).
Solving $x + y = 8$ and $-x + y = 2$ gives $2y = 10 \Rightarrow y = 5$ and $x = 3$.
Solving $x + y = 8$ and $-x + y = -2$ gives $2y = 6 \Rightarrow y = 3$ and $x = 5$.
Thus,the coordinates of $P$ are $(3, 5)$ or $(5, 3)$.