If the image of the point $$(-4, 5)$$ in the line $$x + 2y = 2$$ lies on the circle $$(x + 4)^2 + (y - 3)^2 = r^2$$, then $$r$$ is equal to :

We need to find the image of point $$(-4, 5)$$ in the line $$x + 2y = 2$$, then check if it lies on $$(x+4)^2 + (y-3)^2 = r^2$$.

The formula for the image of point $$(x_1, y_1)$$ in the line $$ax + by + c = 0$$ is:

$$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{-2(ax_1 + by_1 + c)}{a^2 + b^2}$$

Here the line is $$x + 2y - 2 = 0$$, so $$a = 1, b = 2, c = -2$$, and $$(x_1, y_1) = (-4, 5)$$.

$$\frac{x - (-4)}{1} = \frac{y - 5}{2} = \frac{-2((-4) + 2(5) - 2)}{1 + 4} = \frac{-2(4)}{5} = \frac{-8}{5}$$

$$x + 4 = -\frac{8}{5} \Rightarrow x = -4 - \frac{8}{5} = -\frac{28}{5}$$

$$y - 5 = -\frac{16}{5} \Rightarrow y = 5 - \frac{16}{5} = \frac{9}{5}$$

The image is $$\left(-\frac{28}{5}, \frac{9}{5}\right)$$.

Now check if this lies on $$(x+4)^2 + (y-3)^2 = r^2$$:

$$\left(-\frac{28}{5}+4\right)^2 + \left(\frac{9}{5}-3\right)^2 = \left(-\frac{8}{5}\right)^2 + \left(-\frac{6}{5}\right)^2 = \frac{64}{25} + \frac{36}{25} = \frac{100}{25} = 4$$

So $$r^2 = 4$$, which gives $$r = 2$$.

The correct answer is Option 1: 2.