The equation of the image of the circle (x-3)^2 + (y-2)^2 = 1 by the mirror x+y=19 is A) (x-14)^2 + (y-13)^2 = 1 B) (x-15)^2 + (y-14)^2 = 1 C) (x-16)^2 + (y-15)^2 = 1 D) (x-17)^2 + (y-16)^2 = 1"

The correct answer is Option D) $(x-17)^2 + (y-16)^2 = 1$.

To find the image of the given circle under the reflection in the line $x + y = 19$, we need to determine the center of the image circle. The original circle is centered at $(3, 2)$ and has a radius of 1, which remains unchanged under reflection.

Given the line of reflection, $x + y = 19$, and the center of the original circle, $(3, 2)$, the line perpendicular to the mirror line passing through $(3, 2)$ can be used to find the center of the reflected circle. The coordinates of the new center $(\alpha, \beta)$ follow the equations:

$$\frac{\alpha - 3}{1} = \frac{\beta - 2}{1} = -\frac{2(3 + 2 - 19)}{1^2 + 1^2}$$

Evaluating this gives:

$$\alpha - 3 = \beta - 2 = 14$$

Hence, $\alpha = 17$ and $\beta = 16$. Therefore, the equation of the image of the circle after reflection is:

$$(x - 17)^2 + (y - 16)^2 = 1$$

Thus, the answer is Option D).