If the solution curve of the differential equation $$\frac{dy}{dx} = \frac{x+y-2}{x-y}$$ passes through the point (2, 1) and (k+1, 2), k > 0, then

We have the differential equation $$\frac{dy}{dx} = \frac{x + y - 2}{x - y}$$. To simplify, we substitute $$X = x - 1$$ and $$Y = y - 1$$ so that $$x = X + 1$$, $$y = Y + 1$$, and $$\frac{dY}{dX} = \frac{dy}{dx}$$. The equation becomes $$\frac{dY}{dX} = \frac{(X + 1) + (Y + 1) - 2}{(X + 1) - (Y + 1)} = \frac{X + Y}{X - Y}$$.

Now we use the substitution $$Y = vX$$ so that $$\frac{dY}{dX} = v + X\frac{dv}{dX}$$. Substituting gives $$v + X\frac{dv}{dX} = \frac{X + vX}{X - vX} = \frac{1 + v}{1 - v}$$. Hence $$X\frac{dv}{dX} = \frac{1 + v}{1 - v} - v = \frac{1 + v - v + v^2}{1 - v} = \frac{1 + v^2}{1 - v}$$.

Separating variables, we get $$\frac{1 - v}{1 + v^2}\,dv = \frac{dX}{X}$$. We split the left side as $$\frac{1}{1 + v^2}\,dv - \frac{v}{1 + v^2}\,dv = \frac{dX}{X}$$. Integrating both sides gives $$\tan^{-1}(v) - \frac{1}{2}\ln(1 + v^2) = \ln|X| + C$$.

Substituting back $$v = \frac{Y}{X} = \frac{y - 1}{x - 1}$$, we obtain $$\tan^{-1}\!\left(\frac{y-1}{x-1}\right) - \frac{1}{2}\ln\!\left(1 + \frac{(y-1)^2}{(x-1)^2}\right) = \ln|x - 1| + C$$. Now $$\frac{1}{2}\ln\!\left(1 + \frac{(y-1)^2}{(x-1)^2}\right) = \frac{1}{2}\ln\!\left(\frac{(x-1)^2 + (y-1)^2}{(x-1)^2}\right) = \frac{1}{2}\ln\!\left((x-1)^2 + (y-1)^2\right) - \ln|x-1|$$.

Substituting this back, the equation simplifies to $$\tan^{-1}\!\left(\frac{y-1}{x-1}\right) - \frac{1}{2}\ln\!\left((x-1)^2 + (y-1)^2\right) + \ln|x-1| = \ln|x-1| + C$$, which gives $$\tan^{-1}\!\left(\frac{y-1}{x-1}\right) = \frac{1}{2}\ln\!\left((x-1)^2 + (y-1)^2\right) + C$$.

We use the condition that the curve passes through $$(2, 1)$$. Substituting $$x = 2, y = 1$$: $$\tan^{-1}(0) = \frac{1}{2}\ln(1) + C$$, so $$C = 0$$. The solution curve is therefore $$\tan^{-1}\!\left(\frac{y-1}{x-1}\right) = \frac{1}{2}\ln\!\left((x-1)^2 + (y-1)^2\right)$$.

Now we apply the condition that the curve passes through $$(k+1, 2)$$. Substituting $$x = k+1, y = 2$$: $$\tan^{-1}\!\left(\frac{1}{k}\right) = \frac{1}{2}\ln(k^2 + 1)$$. Multiplying both sides by 2 gives $$2\tan^{-1}\!\left(\frac{1}{k}\right) = \ln(k^2 + 1) = \log_e(k^2 + 1)$$.

Hence, the correct answer is Option A.