Let $$\frac{dy}{dx} = \frac{ax - by + a}{bx + cy + a}$$, where $$a, b, c$$ are constants, represent a circle passing through the point $$(2, 5)$$. Then the shortest distance of the point $$(11, 6)$$ from this circle is

We are given $$\frac{dy}{dx} = \frac{ax - by + a}{bx + cy + a}$$ represents a circle passing through $$(2, 5)$$.

For the ODE to represent a circle, we recall that the general equation of a circle can be written as $$F(x, y) = 0$$. Taking the total differential gives $$F_x + F_y \frac{dy}{dx} = 0$$, so $$\frac{dy}{dx} = -\frac{F_x}{F_y}$$. In the case of a circle with equation $$x^2 + y^2 + 2gx + 2fy + d = 0$$, we have $$F_x = 2x + 2g$$ and $$F_y = 2y + 2f$$, which yields $$\frac{dy}{dx} = -\frac{2x + 2g}{2y + 2f} = -\frac{x + g}{y + f}$$.

Next, comparing the given form $$\frac{dy}{dx} = \frac{ax - by + a}{bx + cy + a}$$ with the expression $$-\frac{x + g}{y + f}$$, we see that there must be no $$y$$ term in the numerator and no $$x$$ term in the denominator, so $$b = 0$$. Thus the numerator becomes $$ax + a = a(x + 1)$$ and the denominator becomes $$cy + a$$. Equating these, we write $$\frac{a(x+1)}{cy + a} = -\frac{x + g}{y + f}$$, which implies for some $$\lambda$$, $$a(x+1) = -\lambda(x + g)\quad\text{and}\quad cy + a = \lambda(y + f)\,.$$ From the first equation it follows that $$a = -\lambda$$ and $$a = -\lambda g$$, hence $$g = 1$$. From the second equation we obtain $$c = \lambda$$ and $$a = \lambda f$$, giving $$c = -a$$ and $$f = -1$$.

Substituting these values into the circle equation yields $$x^2 + y^2 + 2(1)x + 2(-1)y + d = 0\,, $$ or equivalently $$x^2 + y^2 + 2x - 2y + d = 0\,. $$ Since the circle passes through $$(2, 5)$$, we substitute to find $$4 + 25 + 4 - 10 + d = 0 \quad\Longrightarrow\quad d = -23\,. $$ Hence the required circle is $$x^2 + y^2 + 2x - 2y - 23 = 0\,. $$

The center of this circle is $$(-1, 1)$$ and its radius is $$\sqrt{1 + 1 + 23} = \sqrt{25} = 5\,. $$ For the point $$(11,6)$$, the distance to the center $$(-1,1)$$ is $$\sqrt{(11-(-1))^2 + (6-1)^2} = \sqrt{144 + 25} = \sqrt{169} = 13\,, $$ so the shortest distance from $$(11,6)$$ to the circle is $$13 - 5 = 8\,. $$

The answer is Option B: $$8$$.