Let the two circles be $S_1: x^2 + y^2 + 2gx + 2fy + c = 0$ and $S_2: x^2 + y^2 + 2g'x + 2f'y + c' = 0$. Their centers are $C_1(-g, -f)$ and $C_2(-g', -f')$. Their radii are $r_1 = \sqrt{g^2+f^2-c}$ and $r_2 = \sqrt{g'^2+f'^2-c'}$, respectively. If two circles cut each other orthogonally, their tangents at the point of intersection are perpendicular. This implies that the radii drawn to the point of intersection are perpendicular. Therefore, the triangle formed by the two centers and a point of intersection (P) is a right-angled triangle with the right angle at P. By the Pythagorean theorem, the square of the distance between the centers equals the sum of the squares of their radii: $(C_1C_2)^2 = r_1^2 + r_2^2$. The distance squared between centers $C_1$ and $C_2$ is: $(C_1C_2)^2 = (-g - (-g'))^2 + (-f - (-f'))^2 = (g' - g)^2 + (f' - f)^2$ $= g'^2 - 2gg' + g^2 + f'^2 - 2ff' + f^2$. The sum of the squares of the radii is: $r_1^2 + r_2^2 = (g^2+f^2-c) + (g'^2+f'^2-c')$. Equating these two expressions: $g'^2 - 2gg' + g^2 + f'^2 - 2ff' + f^2 = g^2+f^2-c + g'^2+f'^2-c'$. Cancelling terms $g^2, f^2, g'^2, f'^2$ from both sides: $-2gg' - 2ff' = -c - c'$. Multiplying by $-1$, we get the condition for orthogonality: $2gg' + 2ff' = c + c'$.

Options A, C, and D do not represent the correct condition for orthogonality. For example, Option A has a difference of constants, while Option C is missing the factor of 2.