If a curve passes through the origin and the slope of the tangent to it at any point $$(x, y)$$ is $$\frac{x^2 - 4x + y + 8}{x - 2}$$, then this curve also passes through the point:

The slope of the tangent at any point $$(x, y)$$ is given by $$\frac{dy}{dx} = \frac{x^2 - 4x + y + 8}{x - 2}$$.

Let $$u = x - 2$$, so $$x = u + 2$$. The equation becomes $$\frac{dy}{du} = \frac{(u+2)^2 - 4(u+2) + y + 8}{u} = \frac{u^2 + 4u + 4 - 4u - 8 + y + 8}{u} = \frac{u^2 + y + 4}{u}$$.

Rewriting: $$\frac{dy}{du} - \frac{y}{u} = u + \frac{4}{u}$$. This is a first-order linear ODE. The integrating factor is $$e^{-\int \frac{du}{u}} = \frac{1}{u}$$.

Multiplying both sides by $$\frac{1}{u}$$: $$\frac{d}{du}\left(\frac{y}{u}\right) = 1 + \frac{4}{u^2}$$.

Integrating: $$\frac{y}{u} = u - \frac{4}{u} + C$$, so $$y = u^2 - 4 + Cu$$.

Substituting back $$u = x - 2$$: $$y = (x-2)^2 - 4 + C(x-2) = x^2 - 4x + C(x-2)$$.

Since the curve passes through the origin $$(0, 0)$$: $$0 = 0 - 0 + C(0-2)$$, giving $$C = 0$$.

Therefore $$y = x^2 - 4x$$. Checking point $$(5, 5)$$: $$25 - 20 = 5$$. This is satisfied.

Therefore, the curve also passes through the point $$(5, 5)$$.