Let R_1 and R_2 be two relations defined on ℝ × ℝ as follows :...

The equivalence class $E_1$ of the point $(4, 0)$ under $R_1$ is given by :
$x^2 + y^2 = 4^2 + 0^2 \Rightarrow x^2 + y^2 = 16$
This represents a circle centered at the origin $(0, 0)$ with radius $r = 4$.

The equivalence class $E_2$ of the point $(a, b)$ under $R_2$ is given by :
$x + y = a + b$
This represents a straight line.

Since the intersection of $E_1$ and $E_2$ contains exactly one element, the line $x + y - (a + b) = 0$ must be tangent to the circle $x^2 + y^2 = 16$.

For tangency, the perpendicular distance from the center of the circle to the line must be equal to the radius of the circle :
$\dfrac{|0 + 0 - (a + b)|}{\sqrt{1^2 + 1^2}} = 4$

$\dfrac{|a + b|}{\sqrt{2}} = 4$

$|a + b| = 4\sqrt{2}$

Squaring both sides, we get :
$(a + b)^2 = (4\sqrt{2})^2 = 32$

Answer: $32$